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11°. 


IN   MEMORIAM 
FLOR1AN  CAJOR1 


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HIGH  SCHOOL  ALGEBRA 


Elementary  Course 


BY 


H.    E.    SLAUGHT,    Ph.D. 

ASSISTANT   PROFESSOR   OF   MATHEMATICS    IN   THE   UNIVERSITY 
OF   CHICAGO 

AND 

N.   J.    LENNES,    M.S. 

INSTRUCTOR   IN    MATHEMATICS   IN   THE   WENDELL   PHILLIPS 
HIGH   SCHOOL,   CHICAGO 


oXKc 


Boston 

ALLYN    and    BACON 
1907 


COPYRIGHT,  1907, 
BY  H.  E.  SLAUGHT 
AND  N.  J.  LENNES. 


5<f3 


PREFACE 

The  High  School  Algebra  has  been  written  in  two  parts,  the 
one  designed  solely  for  the  first  year  and  the  other  for  a  com- 
plete review  and  advanced  course  at  a  later  period. 

The  authors  recognize  the  impossibility  of  combining  in  a 
single  treatment  the  qualities  necessary  for  beginners  with 
the  more  mature  point  of  view  suitable  for  the  third  or  fourth 
year  student. 

The  important  features  of  the  Elementary  Course  are : 

1.  Algebra  is  vitally  and  persistently  connected  with  arithmetic. 

Each  principle  in  the  book  is  first  studied  in  its  applica- 
tion to  numbers  in  the  Arabic  notation.  The  principles  of 
Algebra  are  thus  connected  with  those  already  known  in 
arithmetic.  Letters  are  introduced  as  abbreviations  for  "  a 
number"  or  "any  number."  Literal  expressions  are  called 
number  expressions  —  the  vague  term  "quantity"  is  not  used. 
In  the  exercises,  Arabic  figures  are  constantly  involved  in  the 
same  manner  as  letters.  Checking  by  the  substitution  of 
parilcular  numbers  for  letters  occurs  throughout. 

2.  TJie  principles  of  Algebra  used  in  the  Elementary  Course 
are  enunciated  in  a  small  number  of  short  statements  —  eighteen 
in  all. 

The  purpose  of  these  principles  is  to  furnish  in  simple  form 
a  codification  of  those  operations  of  Algebra  which  are  suffi- 
ciently different  from  the  ones  already  familiar  in  arithmetic 
to  require  special  emphasis.  Such  a  codification  has  several 
important  advantages : 


[5306144 


iv  PREFACE 

By  constant  reference  to  these  few  fundamental  statements 
they  become  an  organic  and  hence  permanent  part  of  the 
learner's  mental  equipment. 

By  their  systematic  use  he  is  made  to  realize  that  the 
processes  of  Algebra  which  seem  so  multifarious  and  hetero- 
geneous are  in  reality  few  and  simple.  Invaluable  training  is 
thus  afforded  in  connecting  things  which  are  essentially,  though 
not  apparently,  related. 

Such  a  body  of  principles  furnishes  a  ready  means  for  the 
correction  of  erroneous  notions,  a  constant  incitement  to  effec- 
tive review,  and  a  definite  basis  upon  which  to  proceed  at  each 
stage  of  progress. 

No  attempt  is  made  at  formal  demonstration  of  these  prin- 
ciples. It  is  believed  that  conviction  as  to  their  validity  is 
most  effectively  brought  to  a  first  year  pupil  by  their  proper 
empirical  exhibition.  Formal  argumentation  is  reserved  for 
the  Advanced  Course. 

3.  TJie  mam  purpose  of  the  Elementary  Course  is  the  solution 
of  problems  rather  than  the  construction  of  a  purely  theoretical 
doctrine  as  an  end  in  itself 

While  the  subject-matter  of  Algebra  is  developed  in  a  logical 
sequence  around  the  principles,  the  attempt  is  made  to  connect 
each  principle  in  a  vital  manner  with  the  learner's  experience 
by  using  it  in  the  solution  of  a  large  number  and  great  variety 
of  simple  problems. 

In  making  the  problems  the  following  criteria  have  been 
observed : 

The  subject-matter  should  be  easily  within  the  comprehen- 
sion of  the  pupil,  and,  so  far  as  possible,  the  problems  should  be 
such  as  one  would  actually  need  to  solve  in  passing  from  known 
to  unknown  data  by  means  of  given  relations.  Such  are  prob- 
lems on  physical  relations  in  Chapter  IV,  and  those  on  geomet- 
rical relations  in  Chapters  VI  and  VII. 


PREFACE  V 

However,  with  the  pupil's  meagre  experience  it  is  impossible 
that  all  problems  should  be  of  this  kind.  Hence  a  considerable 
body  of  problems  has  been  introduced  which  involve  artificial 
relations  imposed  upon  numbers.  Such  problems  are  of  two 
kinds : 

(a)  Those  involving  numbers  related  to  concrete  things,  as, 
for  example,  the  problems  on  pages  44  to  46.  The  data  used  in 
this  class  of  problems  (with  very  few  exceptions)  are  such  as 
have  a  distinct  value  or  real  interest  in  themselves.  Every 
effort  has  been  made  to  have  the  data  entirely  trustworthy. 

(b)  Those  involving  numbers  not  related  to  concrete  things, 
as,  for  example,  problems  12  to  18,  page  43.  Such  problems 
are  used  as  the  basis  for  the  development  of  formulas,  pages 
110  to  114,  Chapter  IV. 

The  utmost  care  has  been  taken  in  grading  the  problems, 
both  as  to  difficulty  and  with  reference  to  the  principles  upon 
which  the  solutions  depend. 

4.  Tlie  order  of  topics  and  the  inclusion  and  exclusion  of 
subject-matter  have  been  determined  by  the  main  purpose  of  the 
Elementary  Course  as  just  stated. 

The  equation,  therefore,  as  the  instrument  for  the  solution 
of  problems,  occurjies  the  leading  -place.  New  topics  are  intro- 
duced only  as  they  are  needed  in  extending  the  use  of  the 
equation.  For  example,  the  whole  subject  of  linear  equations, 
including  those  with  two  and  with  three  unknown  quantities, 
is  treated  before  long  division;  and  quadratic  equatiens  are 
placed  before  the  formal  treatment  of  literal  fractions. 

The  subject  of  factoring  is  introduced  when  quadratic  equa- 
tions are  to  be  solved,  and  is  immediately  used  for  that 
purpose  in  special  cases. 

Long  division  of  polynomials  is  first  found  necessary  as  an 
introduction  to  square  root,  which  together  with  radicals  is 
essential  to  the  solution  of  the  general  quadratic  equation. 


vi  PREFACE 

The  topics  excluded  are :  (a)  complicated  factoring,  (&)  H.  C.F. 
by  the  long  division  process,  (c)  complicated  fractions,  (d)  si- 
multaneous equations  in  more  than  three  unknowns,  (e)  cube 
root,  (/)  fractional  and  negative  exponents,  (g)  equations  con- 
taining complicated  radicals,  Qi)  simultaneous  quadratics  except 
the  case  of  a  quadratic  and  a  linear  equation. 

These  omissions  have  permitted  the  introduction  of  a  much 
greater  number  and  larger  variety  of  problems  than  would 
otherwise  be  possible  without  diminishing  the  number  of  drill 
exercises. 

For  example,  instead  of  an  extended  and  purely  theoretical 
discussion  of  radicals,  only  so  much  is  given  as  is  needed  in 
treating  simple  quadratic  equations;  and  this  is  immediately 
applied  to  the  solution  of  a  body  of  interesting  problems  in- 
volving such  concrete  geometrical  relations  as  are  freely  used 
in  modern  grammar  school  arithmetics. 

This  larger  space  allotted  to  problems  makes  it  possible  to 
pass  by  slow  gradation  from  extremely  simple  to  more  compli- 
cated cases  which  would  otherwise  be  too  difficult,  and  thus  to 
secure  the  facility  and  power  in  interpreting  and  solving  prob- 
lems which  is  demanded  in  physics  and  other  sciences. 

Attention  is  further  called  to  the  following  features : 

The  simple  and  scientific  treatment  of  the  solution  of  the 
equation  as  enunciated  in  Principle  VIII,  page  36. 

The  numerous  illustrative  solutions  which  have  been  given 
in  full  for  the  purpose  of  exhibiting  the  best  methods  of  attack 
and  the  most  effective  arrangement  of  the  work. 

The  development  of  negative  numbers  from  concrete  rela- 
tions, immediately  followed  by  a  large  number  of  problems 
showing  their  practical  use  and  interpretation. 

The  introduction  of  graphs  at  the  beginning  of  the  chapter 
on  simultaneous  equations,  thus  making  the  graph  the  basis 
of  the  study  of  simultaneous  equations. 


PREFACE  Vll 

The  notions  of  a  formula  and  of  substitution  in  a  formula 
are  developed  in  connection  with  subjects  already  familiar, 
such  as  interest,  areas,  volumes,  etc.  (See  page  101.)  These 
notions  are  then  applied  to  the  study  of  other  subjects.  (See 
page  115.) 

Literal  equations  are  in  each  case  introduced  as  a  generaliza- 
tion of  a  series  of  concrete  problems  immediately  preceding. 
(See  page  111.)  Such  equations  are  then  to  be  solved  for  each 
letter  involved.     (See  page  127.) 

Problems  involving  physical  relations  are  in  each  case  intro- 
duced by  means  of  a  series  of  carefully  graded  problems  lead- 
ing up  to  the  general  case.     (See  page  120.) 

If  in  any  case  the  problems  are  found  to  be  too  numerous, 
the  later  sections  of  Chapter  IV  may  be  omitted  or  postponed 
until  after  graphs  and  simultaneous  equations  have  been 
studied.  Many  of  the  exercises  are  simple  enough  to  be  read 
in  class  as  mental  drill  exercises. 

Keview  questions  and  exercises  are  given  at  appropriate 
intervals  throughout  the  book. 

Ratio  and  proportion  are  treated  in  the  chapter  on  fractions 
preparatory  to  their  use  in  plane  geometry. 

The  authors  desire  to  express  their  thanks  to  many  who 
have  given  useful  suggestions,  and  especially  to  Mr.  J.  A. 
Dixon,  Wendell  Phillips  High  School,  Chicago,  and  to  Miss 
Mabel  Sykes,  South  Chicago  High  School,  Chicago,  who  have 
solved  all  the  problems  and  exercises  and  whose  criticisms 
have  been  of  special  value.  Thanks  are  also  due  to  Mr.  Edwin 
Hand,  Jr.,  for  helping  to  prepare  the  cuts. 

H.    E.   SLAUGHT. 
N.   J.   LENNES. 
Chicago,  May,  1907. 


CONTENTS 

CHAPTER  I 
INTRODUCTION  TO  THE  EQUATION 

PAGE 

Peinciple  I.           Addition  of  Numbers  having  a  Common  Factor  .  1 

Peinciple  II.         Subtraction  of  Numbers  having  a  Common  Factor  9 

Peinciple  III.        Multiplication  of  the  Product  of  Several  Factors  12 
Pbinciple  IV.        Multiplication  of  the  Sum  or  Difference  of  Two 

Numbers 14 

Peinciple  V.          Division  of  the  Product  of  Several  Factors  .         .  20 
Peinciple  VI.        Division    of    the    Sum    or    Difference   of    Two 

Numbers 23 

Peinciple  VII.       Number  Expressions  in  Parentheses    .  28 

Peinciple  VIII.     Identities  and  Equations 34 

Directions  for  Written  Work 38 

Equations  involving  Fractional  Coefficients          ...  41 

CHAPTER  II 
POSITIVE  AND  NEGATIVE  NUMBERS 


A  New  Kind  of  Number       .... 
Peinciple  IX.        Addition  of  Signed  Numbers 

Addition  by  Counting 

Averages  of  Signed  Numbers 
Peinciple  X.  Subtraction  of  Signed  Numbers   . 

Peinciple  XL        Multiplication  of  Signed  Numbers 
Peinciple  XII.       Division  of  Signed  Numbers 

Interpretation  and  Use  of  Signed  Numbers  in  Problems 


47 
49 
53 
55 
58 
62 
65 
67 


PAGE 

72 
74 
82 
88 
92 


X  CONTENTS 

CHAPTER  III 

INVOLVED   NUMBER  EXPRESSIONS 

Double  Use  of  the  Signs  +  and  — 
Addition  and  Subtraction  of  Polynomials 
Principle  XIII.     Multiplication  of  Polynomials 
Squares  of  Binomials    .... 
Review  Questions  and  Exercises  . 

CHAPTER  IV 

SOLUTION  OF  PROBLEMS 

Problems  involving  Interest 100 

Problems  involving  Areas 105 

Problems  involving  Volumes 108 

Problems  involving  Simple  Number  Relations      .        .         .  110 

Problems  involving  Motion 115 

Problems  involving  the  Simple  Lever 120 

Problems  involving  Densities 122 

Problems  involving  Momentum 125 

Problems  involving  Thermometer  Readings  .        .        .  128 

Problems  involving  the  Arrangement  and  Value  of  Digits  .  130 

Review  Questions  and  Problems 131 

CHAPTER  V 

INTRODUCTION  TO  SIMULTANEOUS  EQUATIONS 

Graphic  Representation  of  Statistics 138 

Graphic  Representation  of  Motion 142 

Graphic  Representation  of  Equations 146 

Solution  of  Simultaneous  Equations  by  Substitution   .         .  153 
Solution  of  Simultaneous  Equations  by  Addition  or  Sub- 
traction            156 

Problems  involving  Two  Linear  Equations  ....  160 


CONTENTS 


XI 


Linear  Equations  in  Three  Variables 
Problems  involving  Three  Unknowns 
Keview  Questions 


PAGE 

166 
168 
171 


CHAPTER  VI 

SPECIAL  PRODUCTS  AND  FACTORS 

Principle  XIV.     Products  of  Powers  of  the  same  Base 
Principle  XV.      Products  of  Monomials 

Factors  of  Number  Expressions 

Monomial  Factors 

Trinomial  Squares 

The  Difference  of  Two  Squares 

The  Sum  of  Two  Cubes 

The  Difference  of  Two  Cubes 

Trinomials  of  the  Form  x2  +  (a  +  6)  x  +  db 

Trinomials  of  the  Form  ax2  +  bx  +  c 

Factors  found  by  Grouping  . 

Equations  and  Problems  solved  by  Factoring 


172 
175 
177 
178 
179 
183 
185 
186 
188 
191 
193 
197 


CHAPTER  VII 

QUOTIENTS  AND  SQUARE  ROOTS 

Principle  XVI.       Quotients  of  Powers  of  the  same  Base        .         .  210 

Principle  XVII .     Division  of  Monomials       .....  212 

Principle  XVIII.    Square  Roots  of  Monomials        ....  214 

Division  of  Polynomials 216 

Square  Roots  of  Polynomials 221 

Square  Roots  of  Numbers  in  the  Arabic  Notation        .         .  225 

Problems  involving  Square  Roots 234 

Solution  of  Quadratic  Equations  by  Means  of  Square  Root  240 

Quadratic  and  Linear  Equations 246 

Problems  involving  Quadratics 247 

Review  Questions 253 


xii  CONTENTS 

CHAPTER   VIII 
FRACTIONS  WITH  LITERAL  DENOMINATORS 

PAGE 

Common  Factors 255 

Common  Multiples 257 

Reduction  of  Fractions  to  Lowest  Terms     ....  260 

Reduction  of  Fractions  to  a  Common  Denominator     .         .  262 

Addition  and  Subtraction  of  Fractions         ....  267 

Multiplication  and  Division  of  Fractions      ....  270 

Complex  Fractions 277 

Ratio  and  Proportion 279 

Equations  involving  Fractions 283 

Simultaneous  Equations  involving  Fractions        .        .        .  289 

Problems  involving  Fractions       .        .        .        .        .        .  291 


HIGH  SCHOOL  ALGEBRA 

ELEMENTARY  COURSE 

CHAPTER  I 

INTRODUCTION   TO   THE   EQUATION 

ADDITION  OF  NUMBERS  HAVING  A  COMMON  FACTOR 

1.  Algebra  like  arithmetic  deals  with  numbers.  The  num- 
bers of  algebra  include  those  used  in  arithmetic,  and  also  other 
numbers  which  are  defined  as  need  arises  for  them.  The  sym- 
bols employed  to  represent  numbers,  and  the  principles  used 
in  operating  upon  them,  are  introduced  by  means  of  illustra- 
tive problems  as  we  proceed. 

2.  Illustrative  Problem.  The  shortest  railway  route  from 
Chicago  to  New  York  is  912  miles.  How  long  does  it  take  a 
train  averaging  38  miles  an  hour  to  make  the  journey  ? 

Solution  in  words.  The  product  of  the  average  number  of  miles  per 
hour  and  the  required  number  of  hours  equals  the  whole  distance 
traveled.  That  is,  38  multiplied  by  "the  required  number  of  hours  " 
equals  912.  Hence  "  the  required  number  of  hours  "  is  one  thirty- 
eighth  of  912,  or  24. 

Solution  using  abbreviations.  If,  instead  of  the  expression  "the 
required  number  of  hours,"  we  use  the  word  "time,"  or  simply  the 
abbreviation  t,  the  solution  may  be  written : 

38  x  t  =  912. 
Hence  t  =  912  +  38  =  24. 

1 


2  INTRODUCTION  TO   THE  EQUATION 

Illustrative  Problem.  If  in  the  above  problem  a  train  makes 
the  journey  in  18  hours,  find  the  average  number  of  miles  per 
hour. 

Solution.  In  words  we  have,  as  before,  18  multiplied  by  "  the  aver- 
age number  of  miles  per  hour "  equals  912.  Hence  "  the  average 
number  of  miles  per  hour  "  equals  one-eighteenth  of  912,  or  50§. 

Using  for  the  expression  "  the  average  number  of  miles  per  hour," 
the  word  "  rate,"  or  simply  the  abbreviation  r,  the  solution  reads : 

18  x  r  =  912. 

r  *  912  +  18  =  50$. 

It  should  be  clearly  understood  that  t  and  r  represent  num- 
bers whose  values  are  unknown  at  the  outset,  but  which  become 
known  at  the  conclusion  of  the  solution. 

3.  Abbreviations  representing  numbers  and  operations  such  as 
are  used  in  these  problems  occur  constantly  in  algebra.  In 
fact,  the  systematic  use  of  such  abbreviations  is  one  of  the 
chief  distinctions  between  arithmetic  and  algebra. 

4.  The  signs  +,  — ,  x ,  ■+-,  and  =  are  used  in  algebra  with  the 
same  meaning  that  they  have  in  arithmetic.  However,  instead 
of  x  a  point  written  above  the  line  is  often  used.  Thus  2  •  3 
means  2x3. 

The  product  of  two  numbers  represented  by  letters  is  gener- 
ally indicated  by  writing  the  letters  consecutively  with  no  sign 
between  them. 

Thus  rt  means  the  same  as  r  •  t  or  r  x  t.  For  example,  in  such  prob- 
lems as  those  just  given  we  would  write  rt  =  d,  meaning  "  the  number  of 
miles  per  hour  multiplied  by  the  number  of  hours  equals  the  distance 
or  number  of  miles  traveled." 

Similarly  2  r  means  2  •  r,  but  (as  in  arithmetic)  25  means  20  +  5, 
not  2  •  5. 

A  fraction  is  often  used  to  indicate  division 

Thus  ?  =  2-^3;  -  =  r-<. 

3  t 


ADDITION  OF  NUMBERS  3 

PROBLEMS 

In  the  same  manner  as  above  solve  the  following  problems, 
first  by  writing  out  the  solution  in  words  and  then  by  using 
such  abbreviations  as  may  be  found  convenient. 

1.  Five  times  a  certain  number  equals  80.  What  is  the 
number  ?     Use  n  for  the  number. 

2.  Twelve  times  a  number  equals  132.   What  is  the  number  ? 

3.  A  tank  holds  750  gallons.  How  long  will  it  take  a  pipe 
discharging  15  gallons  per  minute  to  fill  the  tank  ? 

4.  The  cost  of  paving  a  block  on  a  certain  street  was  $  7  per 
front  foot.    How  long  was  the  block  if  the  total  cost  was  $  4620? 

5.  A  city  lot  sold  for  $  7500.  What  was  the  frontage  if 
the  selling  price  was  $  225  per  front  foot  ? 

6.  An  encyclopedia  contains  18,000  pages.  How  many 
volumes  are  there  if  they  average  750  pages  to  the  volume  ? 

7.  What  is  the  cost  of  fencing  per  rod  if  it  requires  $  256 
to  fence  a  quarter  section  of  land  ? 

8.  A  square  court  60  feet  on  a  side  is  to  be  paved  with 
square  tiles.  How  many  square  feet  in  each  tile  if  the  pave- 
ment requires  1600  tiles  ? 

9.  An  excavation  for  a  building  is  to  be  130  feet  long, 
80  feet  wide,  and  9  feet  deep.  If  900  cubic  feet  are  removed 
each  day,  how  long  will  it  require  to  complete  it  ? 

10.  A  railway  embankment  which  contains  48,900  cubic 
yards  of  earth  was  completed  in  163  days.  At  what  rate  was 
it  filled  in  ? 

11.  Taking  the  length  of  the  earth's  orbit  as  584  million  miles, 
find  how  far  the  earth  travels  in  one  day ;  also  in  one  hour. 

5.  Definition.  If  a  number  is  the  product  of  two  or  more  num- 
bers, then  these  numbers  are  called  factors  of  the  given  number. 

E.g.  The  integral  factors  of  10  are  1  and  10  or  2  and  5.  2\  and 
4  are  also  factors  of  10. 


4  INTRODUCTION   TO   THE  EQUATION 

6.  Illustrative  Problem.  Divide  the  number  84  into  three 
parts  such  that  the  second  part  is  five  times  the  first  and  the 
third  part  is  eight  times  the  first. 

Solution.  Using  the  abbreviation  p  for  "  the  first  part  of  the 
number,"  we  have       1.porp  =  the  first  part> 

5p  =  the  second  part, 
8  p  =  the  third  part. 

Since  the  sum  of  the  three  parts  is  84, 
p  +  5p  +  &p  =  84. 

To  complete  the  solution  of  this  problem  it  is  necessary  to  find  the 
sum  of  the  numbers  p,  5p,  and  8p  without  knowing  what  number  is 
represented  by  p  itself.     If  we  suppose  this  sum  to  be  lip,  then 

Up  a  84 
and  p  =  84  +■  14  =  6. 

This  supposition  leads  to  the  correct  result  since,  if  p  =  6,  then 
/>+5j9  +  8jp  =  6  +  5.6  +  8.6=6  +  30  +  48=84.  Hence  the  three 
parts  into  which  84  is  divided  are  6,  30,  and  48. 

7.  The  process  here  used  for  adding  p,  5p,  and  8p  is  a  new 
method  of  adding  which  is  of  very  great  importance  in 
algebra.  This  method  is  further  exhibited  by  the  following 
examples : 

(1)  To  add  18,  42,  54,  and  30,  we  first  factor  these  numbers  so  as 
to  show  the  common  factor  6  and  then  add  the  remaining  factors  3,  7, 
9,  and  5,  and  multiply  this  sum  by  the  common  factor  6.  The  result 
is  24 •  6,  or  144.     The  work  may  be  arranged  as  follows: 

18  a  3-6  Similarly  (2) 

42=  7-6 

54=  9-6 

30=  5-6 


16  = 

1 

16  = 

2 

8  = 

4 

1 

64  = 

4 

16  = 

8 

8  = 

16 

1 

32  = 

2 

16  = 

4 

8  = 

8 

4 

48  = 

3 

16  = 

6 

8  = 

12 

4 

160  = 

10 

16  = 

20 

8  = 

40 

i 

144  =  24  .  6 

From  example  (2)  we  see  that  in  case  the  numbers  to  be 
added  have  two  or  more  common  factors  it  does  not  matter 


ADDITION  OF  NUMBERS  5 

which  one  is  selected.  If  each  number  is  separated  into  tivo 
factors  one  of  which  is  a  common  factor,  then  it  is  evident 
that  the  sum  in  any  case  is  found  by  multiplying  the  common 
factor  by  the  sum  of  the  other  factors. 

In  the  above  manner  add  the  following  sets  of  numbers  and 
show  in  each  case  by  adding  in  the  ordinary  way  that  the  result 
is  correct : 


14 

36 

32 

17 

39 

32 

28 

72 

64 

34 

52 

48 

35 

48 

128 

68 

78 

40 

70 

12 

256 

85 

91 

72 

8.  Definition.  If  a  number  is  the  product  of  two  factors, 
then  either  of  these  factors  is  called  the  coefficient  of  the  other 
in  that  number. 

E.g.  In  2  •  3,  2  is  the  coefficient  of  3,  and  3  is  the  coefficient  of  2. 
In  9  rt,  9  is  the  coefficient  of  rt,  r  is  the  coefficient  of  9 t,  and  t  is  the 
coefficient  of  9  r.  In  such  expressions  as  9  rt  the  factor  represented  by 
Arabic  figures  is  usually  regarded  as  the  coefficient. 

The  preceding  examples  illustrate  the  following  principle : 

9.  Principle  I.  To  add  numbers  having  a  common 
factor,  add  the  coefficients  of  the  common  factor  and 
multiply  tJie  sum  by  the  common  factor. 

If  n  represents  some  number,  then  4»i,  in  the  same  dis- 
cussion, means  four  times  that  number,  and  5  n  means  five 
times  that  number.  Hence,  by  Principle  I,  4  n  -f-  5  n  =  9  n, 
which  is  true  no  matter  what  number  is  represented  by  n. 

By  means  of  this  principle  perform  the  following  additions, 
understanding  that  each  letter  represents  some  number : 

1.  Sx  +  7  a +  16  a +2  a.  4.  7b+8b  +&  +  6&. 

2.  13 n  +  8 n  +  7 n  +  9 n.  5.  8 £  +  7 t  +  5t. 

3.  3a  +  a  +  2 a  +  6 a.  6.  3r  +  5r  +  llr. 


6  INTRODUCTION   TO   THE  EQUATION 

Show  the  correctness  of  the  result  in  each  of  the  above  by 
letting  x  =  2,  n  =  1,  a  =  4,  6  =  3,  t  =  6,  and  r  =  7.  Try  also 
other  values  for  the  letters. 

Such  a  test  for  the  correctness  of  an  operation  is  called  a  check. 

10.  Definitions.  Combinations  of  Arabic  figures  or  letters,  or 
both,  by  means  of  the  signs  of  operation,  +,  — ,  etc.,  are  called 
number  expressions. 

E.g.  38,  18  r,  p  +  5 p  +  8p,  are  number  expressions. 

Two  number  expressions  representing  the  same  number,  when 
connected  by  the  sign  =,  form  an  equality. 

The  expressions  thus  connected  are  called  the  members  of  the 
equality  and  are  distinguished  as  the  right  and  left  members. 

Equalities,  such  as  8t-\-7 1  =  15t,  in  which  the  letters  may 
be  any  numbers  whatever,  are  called  identities.  Equalities  of 
the  type  p  +  5p  +  8p  =  84  are  called  equations.    (See  §  §  29-34.) 

EXERCISES 

1.  Add  5 1,  11 t,  20 t,  and  47 t.  Check  the  result  by  letting 
t  =  3 ;  also  t  =  50,  and  t  =  150. 

2.  Add  7  r,  23  r,  28  r,  52  r,  and  117  r.  Check  the  result  for 
r  =  11,  r  =  20,  and  r  =  1. 

3.  Add  3  rt,  7  rt,  65  rt,  and  16  rt.     Check  for  r  =  1,  t  =  2. 

4.  Add  1^  w,  2|  w,  3^  n,  and  check  for  n  =  6. 

5.  Add  66,  88,  99,  and  121  by  Principle  I. 

6.  Add  144,  96, 120,  and  50  •  12  by  Principle  I. 

7.  Add  5  •  40,  8  .  60,  and  6  •  20  by  Principle  I. 

8.  Add  5  ax,  3  ax,  and  7  ax.     Check  for  a  =  2,  x  =  4. 

9.  Add  7  ax,  3  bx,  and  12  ex. 

In  this  case  the  common  factor  is  x.     Hence  using  Principle  I, 

7  ax  +  3  bx  +  12  ex  =  (7  a  +  3  6  + 12  c)x.  The  parenthesis  here  indicates 
that  the  numbers  represented  by  x  and  the  expression  7  a  +  3  b  +  12  c 
are  to  be  multiplied. 


ADDITION  OF  NUMBERS 


SOLUTION  OF  PROBLEMS 


11.  One  great  object  in  the  study  of  algebra  is  to  simplify 
the  solution  of  problems.  This  is  done  by  using  letters  to  repre- 
sent the  unknown  numbers,  by  stating  the  problem  in  the  form 
of  an  equation,  and  by  arranging  the  successive  steps  of  the 
solution  in  an  orderly  manner. 

Skill  in  translating  problems  into  equations  depends  upon 
attention  to  the  following  points  : 

(1)  Read  and  understand  clearly  the  statement  of  the  prob- 
lem, as  it  is  given  in  words. 

(2)  Select  the  unknown  number,  and  represent  it  by  a  suit- 
able letter,  say  the  initial  letter  of  a  word  which  will  keep  its 
meaning  in  mind.  If  there  are  more  unknown  numbers  than 
one,  try  to  express  the  others  in  terms  of  the  one  first  selected. 

(3)  Find  two  number  expressions  which,  according  to  the 
problem,  represent  the  same  number,  and  set  them  equal  to 
each  other,  thus  forming  an  equation. 

These  steps  are  exhibited  in  the  following  solution : 

A  tree  108  feet  high  was  broken  off  by  the  wind  so  that  the 

part  left  above  the  first  branch  was  three  times  as  long  as  the 

part  broken  off,  and  the  part  below  the  first  branch  was  twice  as 

long  as  the  part  broken  off.    How  long  was  the  part  broken  off  ? 

Solution.     Let  6  represent  the  number  of  feet  broken  off. 

Then      36  is  the  number  of  feet  left  above  the  first  branch, 
and  2  6  is  the  number  of  feet  below  the  first  branch. 

Hence,  b  +  36  +  26  and  108  are  number  expressions,  each  repre- 
senting the  total  height  of  the  tree. 

Therefore  6  +  36  +  26  =  108.  (1) 

By  Principle  I,  66  =  108.  (2) 

Then  6  =  one  sixth  of  108,  or  18.  (3) 

Hence,  the  part  broken  off  was  18  feet  long. 

Equation  (3)  is  derived  from  (2)  by  dividing  both  members  by  6. 


8  INTRODUCTION  TO   THE  EQUATION 

PROBLEMS 

1.  The  greater  of  two  numbers  is  5  times  the  less,  and 
their  sum  is  180.     What  are  the  numbers  ? 

2.  A  number  increased  by  twice  itself,  4  times  itself, 
and  6  times  itself,  becomes  429.     What  is  the  number  ? 

3.  A  father  is  3  times  as  old  as  his  son,  and  the  sum  of 
their  ages  is  48  years.     How  old  is  each  ? 

4.  In  a  company  there  are  39  persons.  The  number  of 
children  is  twice  the  number  of  grown  people.  How  many 
are  there  of  each  ? 

5.  A  and  B  receive  $45  for  doing  a  certain  piece  of  work. 
If  A  gets  4  times  as  much  as  B,  how  much  does  each  receive  ? 

6.  The  population  of  Tokio  is  twice  that  of  Canton,  and  the 
sum  of  their  populations  is  2,700,000.  How  many  inhabitants 
in  each  city  ? 

7.  Find  two  consecutive  integers  whose  sum  is  133. 

8.  The  area  of  Louisiana  is  (nearly)  4  times  that  of 
Maryland,  and  the  sum  of  their  areas  is  60,930  square  miles. 
Find  the  (approximate)  area  of  each  state. 

9.  The  horse-power  of  a  certain  steam  yacht  is  12  times 
that  of  a  motor  boat.  The  sum  of  their  horse-powers  is  195. 
Find  the  horse-power  of  each. 

10.  There  are  three  circles  on  the  blackboard.  The  circum- 
ference of  the  second  is  5  times  that  of  the  first,  and  the 
circumference  of  the  third  is  10  "times  that  of  the  first.  The 
sum  of  their  circumferences  is  16  feet.  Find  the  circumfer- 
ence of  each. 

11.  At  a  football  game  there  were  2000  persons.  The  num- 
ber of  women  was  3  times  the  number  of  children,  and  the 
number  of  men  was  6  times  the  number  of  children.  How 
many  men,  women,  and  children  were  there  ? 


SUBTRACTION  OF  NUMBERS  9 

12.  The  population  of  Portland,  Oregon  (estimate  of  the 
Census  Bureau,  1904),  was  twice  that  of  Dallas,  Texas,  and  the 
population  of  Toledo  was  3  times  that  of  Dallas.  The 
three  cities  together  had  300  thousand  inhabitants.  How 
many  were  there  in  each  city? 

Let  >i  =  number  of  thousands  of  inhabitants  in  Dallas. 
Then    2ra  =  number  of  thousands  of  inhabitants  in  Portland, 
and     3  n  =  number  of  thousands  of  inhabitants  in  Toledo. 
Hence  n  +  2n  +  3n  =  300. 

13.  It  is  twice  as  far  from  New  York  to  Syracuse  as  from 
New  York  to  Albany,  and  it  is  4  times  as  far  from  New 
York  to  Cleveland  as  from  New  York  to  Albany.  The  sum  of 
the  three  distances  is  1015  miles.     Find  each  distance. 

14.  In  Maryland  there  were  (census  of  1900)  4  times  as 
many  whites  as  negroes.  The  total  population  was  1185  thou- 
sand.    How  many  of  each  were  there  ? 

If  n  equals  the  number  of  thousands  of  negroes,  then  the  equation 
is  n  +  in  =  1185. 

SUBTRACTION  OF  NUMBERS  HAVING  A  COMMON  FACTOR 
12.   Numbers  having  a  common  factor  may  be  subtracted  in 
a  manner  similar  to  the  process  exhibited  under  Principle  I. 

Thus,  from  64  =  8  •  8  From  84  =  12  -  7  From  17  n 

subtract       48  =  6-8         subtract  49  =    7-7        subtract    6n 
Eemainder  16  =  2  . 8  35=    5-7  liw 

In  like  manner  perform  the  following  subtractions : 

1.  9.7-3-7.  5.   6.99-5.99.  9.    Qn-2n. 

2.  10-4-6-4.         6.   20-19-13-19.        10.   6-50-2-50. 

3.  8.8-2-8.  7.    8a  -3a.  11.    106-4 6. 

4.  5-11-3-11.       8.   8-5-3-5.  12.   7a -4a. 


10  INTRODUCTION    TO    THE    EQUATION 

These  examples  illustrate  the  following  principle : 

13.  Principle  II.  To  find  the  difference  of  two  numbers 
having  a  common  factor,  subtract  the  coefficients  of  the 
common  factor  and  multiply  the  result  by  tlie  common 
factor. 

Illustrative  Problem.  If  thirteen  times  a  certain  number 
diminished  by  eight  times  the  number  equals  75,  what  is  the 
number  ? 

Solution.     Let  n  represent  the  required  number. 
Then   13  n  —  8  n  and  75  are  expressions  representing  the  same 
number. 

Hence,  13  n  —  8  n  =75. 

By  Principle  II,  5n  =  75. 

Dividing  each  member  by  5,      n  =15,  the  required  number. 

Check.  13  •  15  -  8  •  15  =  195  -  120  =  75. 


EXERCISES  AND  PROBLEMS 

1.  By  means  of  Principle  II  subtract  72  from  160 ; 
50  from  300;  39  from  78;  34  from  85;  58  from  174;  and 
69  from  161. 

2.  Subtract  109  .  87  from  209  •  87  by  Principle  II. 
Check  by  first  finding  the  products  and  then  subtracting  as  in 
arithmetic. 

Perform  the  following  indicated  operations  and  check 
those  in  which  letters  are  involved  by  substituting  convenient 
numbers  : 

3.  6St-llt.  7.   3.4n+5.4w+11.4w-7-4n. 

4.  15w+25w-18n.  8.   13rt+16rt+3rt-20rt. 

5.  70aj-15»+7aj-23a?.        9.   144-96+50.12-20-12. 

6.  18- 7-3- 7- 2- 7+6- 7.  10.   lla«-3aa;+4aa;. 


SUBTRACTION  OF  NUMBERS  11 

11.  11  ax  —  dbx  +  kcx.  19.    ar  +  br  —  cr. 

12.  11- 9 -6 -9 +  3 -9.  20.   3ry-2sy-ty. 

13.  20»-6n  +  2w.  21.    11.17  +  47-17-8.17. 

14.  an  —  bn  +  en.  22.    accy  +  bxy  —  3  ay. 

15.  5*  +  2(H-3£.  23.    3a&c  +  7a6c-2a&c. 

16.  8s-3s  +  20s.  24.    7  -5  a -3  -5^  +  8.  5  x. 

17.  6a-4a  +  3a-2a.  25.    a«2r  +  6-2r  —  c-2r. 

18.  llrs  — 2rs  +  4rs.  26.    2ar  +  26r  —  2cr. 

27.  Four  times  a  certain  number  plus  3  times  the  number 
minus  5  times  the  number  equals  48.     What  is  the  number  ? 

28.  One  number  is  4  times  another,  and  their  difference  is  9. 
What  are  the  numbers  ? 

29.  Find  a  number  such  that  when  4  times  the  number  is 
subtracted  from  12  times  the  number  the  remainder  is  496. 

30.  The  population  of  Ohio  (1901)  was  twice  that  of  Wis- 
consin. The  difference  of  their  populations  was  2100  thou- 
sand.   Find  the  population  of  each  state. 

31.  The  population  of  Illinois  in  1903  was  5  times  as  great 
as  that  of  West  Virginia.  The  difference  between  their  popu- 
lations was  4080  thousand.     What  was  the  population  of  each  ? 

32.  There  are  three  numbers  such  that  the  second  is  11  times 
the  first  and  the  third  is  27  times  the  first.  The  difference 
between  the  second  and  the  third  is  64.     Find  the  numbers. 

33.  A  cubic  foot  of  asphaltum  is  twice  as  heavy  as  a  cubic 
foot  of  light  anthracite  coal.  Seven  cubic  feet  of  coal  weigh  390 
pounds  more  than  1  cubic  foot  of  asphaltum.  Find  the  weight 
per  cubic  foot  of  each. 

34.  Thirty-nine  times  a  certain  number,  plus  19  times  the 
number,  minus  56  times  the  number,  plus  22  times  the  num- 
ber, equals  12.     Find  the  number. 


12  INTRODUCTION  TO   THE  EQUATION 

MULTIPLICATION  OF  A  PRODUCT 

14.  Illustrative  Problem.  Three  men,  A,  B,  and  C,  invest 
together  $  33,000.  B  puts  in  twice  as  much  as  A,  and  C  4 
times  as  much  as  B.     How  much  does  each  invest  ? 

Solution.  Let  d  represent  the  number  of  dollars  invested  by  A. 
Then  2  d  represents  B's  investment,  and  4  (2  d)  represents  C's  invest- 
ment. 

Hence,  d  +  2d  +  4  (2d)=  33000. 

To  complete  the  solution  of  this  problem  it  is  necessary  to  multiply 
2  c?  by  4  without  knowing  what  number  is  represented  by  d.    If  we 
suppose  the  product  to  be  8  d, 
then  d  +  2d  +  8d  =  33000. 

By  Principle  I,  1 1  d  =  33000. 

Dividing  each  member  by  11,  d  =    3000,  A's  investment ; 
2c?=    6000,  B's  investment ; 
4-26?  =  24000,  C's  investment. 

The  supposition  that  4  (2  d)  =  8  d  is  justified  by  the  fact  that  the 
numbers  thus  found  satisfy  the  conditions  of  the  problem. 

That  is,     6?+2tf+4(2o?)=  3000  +  6000  +  24000  =  33000. 

15.  The  process  here  used  for  multiplying  2  d  by  4  is  of  great 
importance  in  algebra. 

The  following  examples  further  exhibit  this  new  method  of 
multiplying : 

1.   4(3- 5)  =4. 15  =  60  2.   2(3-4.5)  =  2-60      =120 

Also  4(3-5)  =  12- 5  =  60  Also  2(3  •  4  .  5)=6  -4  -5   =120 

And  4(3  •  5)  =  3  •  20  =  60         And  2(3  •  4  •  5)  =  3  •  8  -  5   =120 

And  2(3 -4- 5)=  3- 4- 10  =  120 

In  like  manner  find  the  following  products  in  two  or  more 
ways  : 

3.  6(2-3).       5.   2(5-149).       7.   20(5  a- 4).      9.   8(4*.  3). 

4.  4(25-99).  6.   4(19-5).         8.    16  (4  s.  7).     10.   9(xy-5). 


MULTIPLICATION  OF  A   PRODUCT  13 

These  examples  illustrate  the  following  principle : 

16.  Principle  III.  To  multiply  the  product  of  several 
factors  by  a  given  number,  multiply  any  one  of  the  factors 
by  that  number,  leaving  the  others  unchanged. 

EXERCISES  AND  PROBLEMS 

Multiply  as  many  as  possible  of  the  following  in  two  or 
more  ways.     Check  where  letters  are  involved. 

7.   4(19-25).  13.   2(rs-16). 

8. '5  (17- 20 -3).  14.   5(xy-5z). 

9.    15(7a&).  15.   7(3-4a&). 

10.  3 (4 ran).  16.   9(a-5-xy). 

11.  5(abc).  17.   4(125-17). 

12.  7(2xy).  18.   40(25-29). 

19.  There  are  three  numbers  whose  sum  is  80.  The  second 
is  3  times  the  first  and  the  third  is  twice  the  second.  What  are 
the  numbers  ? 

20.  There  are  three  numbers  such  that  the  second  is  8 
times  the  first  and  the  third  is  3  times  the  second.  If  the 
second  is  subtracted  from  the  third  the  remainder  is  48.  Find 
the  numbers. 

21.  The  population  of  Bridgeport,  Connecticut,  is  twice  that 
of  Butte,  Montana.  Three  times  the  population  of  Bridgeport 
plus  twice  that  of  Butte  equals  320  thousand.  Find  the  popu- 
lation of  each  city. 

22.  It  is  4  times  as  far  from  New  York  City  to  Cincinnati 
as  from  New  York  to  Baltimore.  Twice  the  distance  from  New 
York  to  Cincinnati  minus  5  times  that  from  New  York  to  Bal- 
timore equals  567  miles.  How  far  is  it  from  New  York  to  each 
of  the  other  cities  ? 


1. 

7(3-4-5). 

2. 

8(7-2-3). 

3. 

9(2-3-4). 

4. 

5(2  ab). 

5. 

3(5  xy). 

6. 

12(8-4-20) 

14  INTRODUCTION   TO   THE  EQUATION 

23.  The  population  of  Hartford,  Connecticut,  is  3  times  that 
of  Oshkosh,  Wisconsin.  Four  times  the  population  of  Hart- 
ford plus  5  times  that  of  Oshkosh  equals  510  thousand.  Find 
the  population  of  each  city. 

24.  One  cubic  inch  of  emery  weighs  twice  as  much  as  1 
cubic  inch  of  ivory.  The  combined  weight  of  10  cubic  inches 
of  each  substance  is  2.1  pounds.  Find  the  weight  per  cubic 
inch  of  each. 

25.  It  is  twice  as  far  from  Boston  to  Quebec  as  from  Boston 
to  Albany  and  3  times  as  far  from  Boston  to  Jacksonville, 
Florida,  as  from  Boston  to  Quebec.  How  far  is  it  from  Boston 
to  each  of  the  other  three  cities,  the  sum  of  the  distances  being 
1818  miles  ? 

26.  A  cubic  inch  of  porcelain  china  is  twice  as  heavy  as 
a  cubic  inch  of  ebony,  and  a  cubic  inch  of  rolled  zinc  is. 3 
times  as  heavy  as  a  cubic  inch  of  porcelain.  The  combined 
weight  of  1  cubic  inch  of  each  substance  is  .387  pounds.  Find 
the  weight  per  cubic  inch  of  each. 

MULTIPLICATION  OF  THE  SUM   OR  DIFFERENCE   OF  TWO  NUMBERS 

17.  Illustrative  Problem.  The  length  and  width  of  a  rectan- 
gle together  equal  58  inches.  If  the  width  were  5  inches 
greater,  the  length  of  the  rectangle  would  then  be  twice  its 
width.     Find  its  dimensions. 

Solution.     Let  10  represent  the  number  of  inches  in  the  width.     If 
it  were  5  inches  wider,  it  would  then  be  w  +  5  inches.     Hence  the 
<  length  is  2  (w  +  5),  the  parenthesis  indicating  that  the  sum  of  w  and  5 
is  to  be  found  and  the  result  multiplied  by  2. 
Hence  the  sum  of  the  length  and  width  is 

id  +  2  (to  +  5)  =  58. 
The  solution  of  this  problem  involves  the  multiplication  of  the 
number  (w  +  5)  by  2  without  knowing  what  number  is  represented 
by  w. 


MULTIPLYING  A    SUM   OR  DIFFERENCE  15 

If  we  suppose  that  2  (w  +  5)  =  2  w  +  10, 

then  we  have         w  +  2  w  +  10  =  58.  (1) 

By  Principle  I,  3  w  +  10  =  58.  (2) 

Hence  3w  =  48,  since  58  is  10  more  than  3  w,  (3) 

and  w  =  16,  the  width.  (4) 

Then  58  -  16  =42,  the  length. 

The  correctness  of  the  above  process  is  shown  by  the  fact  that  the 
numbers  42  and  16  fulfill  the  conditions  of  the  problem ; 
that  is,  2  (16  +  5)  =  2  .  21  =  42, 

and  16  +  42  =  58. 

Equation  (3)  is  derived  from  (2)  by  subtracting  10  from  each 
member. 

18.  The  multiplication  of  the  number  (w  +  5)  by  2  without 
knowing  in  advance  what  number  is  represented  by  w  is  a  new 
method  of  multiplying  which  is  constantly  used  in  algebra.  The 
following  examples  further  exhibit  this  method : 

(1)  4(2  +  7)  =  4-9  =  36, 

or  4(2  +  7)=4- 2  +  4- 7  =  8  +  28  =  36. 

(2)  3(3  +  8  +  9)  =  3- 20  =  60, 

or  3(3  +  8 +  9)  =3- 3 +  3 -8 +3- 9  =  9  +  24  +  27  =  60. 

It  is  thus  seen  that  in  each  case  the  same  result  is  obtained 
whether  we  first  add  the  numbers  in  the  parenthesis  and  then 
multiply  the  sum  or  first  multiply  the  numbers  in  the  parenthe- 
sis one  by  one  and  then  add  the  products. 

Multiply  each  of  the  following  in  two  ways  where  possible : 

1.  3(2  +  7).  5.  3(a  +  6).  9.  a(3  +  7  +  10). 

2.  5(3  +  4  +  5).  6.  ll(h  +  k).  10.  15(x  +  y  +  z). 

3.  8(5  +  9+7).  7.  4(5a  +  76  +  c).       11.  20(ra+w+_p). 

4.  7(6+11+9+9).       8.  a(5  +  4  +  7).  12.  9(2r+3s+*). 


16  INTRODUCTION   TO   THE  EQUATION 

19.  In  a  manner  similar  to  the  above  the  difference  of  two 
numbers  represented  by  Arabic  figures  may  be  multiplied  by  a 
given  number  in  either  of  two  ways.  ■ 

E.g.         6(8-3)  =  6 -5  =  30, 
or  6(8-3)  =  6-8-6-  3  =  48-18  =  30. 

The  same  result  is  obtained  whether  we  first  perform  the  sub- 
traction indicated  in  the  parenthesis  and  then  multiply  the  dif- 
ference, or  first  multiply  the  numbers  separately  and  then 
subtract  the  products.  In  the  case  of  numbers  represented  by 
letters  evidently  the  second  process  only  is  available. 

E.g.      6(r-t)  =  6r-6t. 

Perforin  as  many  as  possible  of  the  following  multiplications 
in  two  ways : 

1.  7(9-2).      4.   17(18-11).    7.   5(3-1).     10.   m{r-s). 

2.  12(17-7).  5.   9 (a -2).  8.   3(y-2).     11.   x{y-z). 

3.  5(12-8).    6.   8  (ft -4).  9.   a(c-d).    12.    t(u~v). 

The  second  of  the  above  methods  is  needed  in  problems  like 
the  following : 

Illustrative  Problem.  The  rate  of  an  express  train  plus  that 
of  a  freight  is  70  miles  per  hour.  If  the  rate  of  the  freight 
were  7  miles  less,  the  express  would  be  going  twice  as  fast  as 
the  freight.     Find  the  rate  of  each. 

Solution.     Suppose  the  rate  of  the  freight  is  now  r  miles  per  hour. 

Then  r—  7  is  the  supposed  rate  of  the  freight, 

and  2  (r  —  7)  is  the  present  rate  of  the  express. 

Hence  r  +  2  (r  -  7)  =  70,  the  sum  of  the  rates,  (1) 

and  r+  2r-  14  =  70.  (2) 

By  Principle  I,  3  r  -  14  =  70.  (3) 


MULTIPLYING  A   SUM  OR  DIFFERENCE  17 

Then  3  r  =  84,  since  70  is  14  less  than  3  r.     (4) 

Hence  r  =  28,  the  rate  of  the  freight, 

and  2  (28  -  7  )  =  2  •  21  =  42,  the  rate  of  the  express. 
Check.  42  +  28  =  70. 

Equation  (4)  is  derived  from  (3)  by  adding  14  to  both  members 
and  observing  that  14  —  14  =  0. 

The  foregoing  examples  illustrate  the  following  principle  : 

20.  Principle  IV.  To  multiply  the  sum  or  difference  of 
two  numbers  by  a  given  number,  multiply  each  of  the 
numbers  separately  by  the  given  number,  and  add  or  sub- 
tract the  products. 

21.  Principles  III  and  IV  should  be  carefully  contrasted,  as 
in  the  following  example : 

2  (2  •  3  •  5).=  4- 3- 5  =  2- 6- 5  =  2- 3- 10, 

but  2(2  +  3 +  5)  =4  +  6  +  10. 

In  multiplying  the  product  of  several  numbers  we  operate 
upon  any  one  of  them,  but  in  multiplying  the  sum  or  difference 
of  numbers  we  operate  upon  each  of  them. 

EXERCISES  AND  PROBLEMS 

1.  Multiply  5  +  7  +  11  by  3  without  first  adding,  and  then 
check  by  performing  the  addition  before  multiplying. 

2.  Multiply  m  +  n  by  4  and  check  for  m  =  5,  n  =  7. 

4  (m  +  n)  =  4  m  +  4  n 
Check.  4  (5  +  7)  =  4  •  12  =  48,  also 

4  •  5  +  4  •  7  =  20  +  28  =  48. 

3.  Multiply  r  +  s  +  x  by  a  and  check  for  r  ==s  =  x  =  a  =  2. 

4.  Multiply  a  +  6  +  c  by  m  and  check  for  a  =  1,  b  =  3, 
c  =  5,  m  =  4. 

5.  Multiply  x  +  y  by  r  and  check  for  x  =  2,  y  =  4,  r  =  6. 


18  INTRODUCTION   TO   THE  EQUATION 

Where  letters  are  involved,  check  the  results  in  the  follow- 
ing by  substituting  convenient  values  : 

6.  8(13-5).  16.   8(2a-3&+4c). 

7.  71(12  +  41-36).  17.   37(3x-2y-z). 

8.  9(o-6  +  8).  18.    13(5r-3x  +  t). 

9.  3(a  +  x  +  y  -11).  19.   20(2a  +  3x  -  5y). 

10.  ra  (a  +  &  —  c).  20.  7  (11  s  —  2  s  +  3 t). 

11.  a(18-7).  21.  35(x-2y  +  3z). 

12.  2  (13  +  8  +  9  -  21).  22.  78  (10  m  +  lln  +  2  r). 

13.  7(3 +  8 +  9 -a).  23.  4(25a;  +  32a;-2/). 

14.  5(17+ a -6).  24.  3(13 a?  +  14 y  -  t). 

15.  32 (a?—  y-M).  25.  7(4a-36  +  c). 

26.  Find  two  consecutive  integers  such  that  3  times  the  first 
plus  7  times  the  second  equals  217. 

27.  Find  two  consecutive  integers  such  that  7  times  the  first 
plus  4  times  the  second  equals  664. 

28.  There  are  three  numbers  such  that  the  second  is  17  less 
than  the  first,  and  the  third  is  8  times  the  second.  The  sum  of 
the  first  and  third  is  89.     What  are  the  numbers  ? 

29.  The  number  of  representatives  and  senators  together  in 
the  United  States  Congress  is  476.  The  number  of  represen- 
tatives is  26  more  than  4  times  the  number  of  senators.  Find 
the  number  of  each. 

30.  The  area  of  Illinois  is  6750  square  miles  more  than  10 
times  that  of  Connecticut.  The  sum  of  their  areas  is  61,640 
square  miles.     Find  the  area  of  each  state. 

31.  The  sum  of  the  horse-powers  of  the  steamships  Campania 
and  Mauritania  is  102  thousand.  The  Mauritania  has  12 
thousand  horse-power  more  than  twice  that  of  the  Campania. 
What  is  the  horse-power  of  each  ship? 


MULTIPLYING  A   SUM  OB    DIFFERENCE  19 

32.  The  population  of  Wyoming  (census  of  1900)  was  50 
thousand  more  than  that  of  Nevada,  and  the  population  of 
Utah  was  93  thousand  more  than  twice  that  of  Wyoming. 
The  population  of  Utah  was  277  thousand.  Find  the  popula- 
tion of  Nevada  and  Wyoming. 

33.  It  is  73  miles  farther  from  Newark  to  Philadelphia  than 
from  New  York  to  Newark,  and  it  is  one  mile  more  than  10 
times  as  far  from  Philadelphia  to  Chicago  as  from  Newark  to 
Philadelphia.  The  sum  of  the  distances  from  New  York  to 
Newark  and  from  Philadelphia  to  Chicago  is  830  miles.  Pind 
each  of  the  three  distances,  and  the  total  distance  from  New 
York  to  Chicago. 

Let      .  d  =  number  of  miles  from  New  York  to  Newark. 

Then  d  +  73  =  number  of  miles  from  Newark  to  Philadelphia, 

and   10  (d  +  73)  +  1  =  number  of  miles  from  Philadelphia  to  Chicago. 
Hence  d  +  10  (d  +  73)  +  1  =  830. 

34.  In  the  championship  season  of  1906  the  Chicago  National 
League  baseball  team  lost  20  games  less  than  New  York,  and 
Pittsburg  lost  12  less  than  twice  as  many  games  as  Chicago. 
Pittsburg  and  New  York  together  lost  116  games.  How  many 
did  each  of  the  three  teams  lose? 

35.  Pikes  Peak  is  3282  feet  higher  than  Mt.  ./Etna,  and 
Mt.  Everest  is  708  feet  more  than  twice  as  high  as  Pikes  Peak. 
The  sum  of  the  altitudes  of  Mt.  iEtna  and  Mt.  Everest  is 
39,867  feet.     Find  the  altitude  of  each  of  the  three  mountains. 

36.  The  altitude  of  Chimborazo  is  22,820  feet  less  than 
3  times  that  of  Mt.  Shasta.  What  is  the  altitude  of  each 
mountain  if  Chimborazo  is  6060  feet  higher  than  Mt.  Shasta  ? 

Let  x  =  the  number  of  feet  in  the  altitude  of  Mt.  Shasta.  Then 
3x  -22820  =  altitude  of  Chimborazo,  and  3a;  -  22820  -  x  =  6060. 

37.  The  distance  of  Mars  from  the  sun  is  39  million 
miles  less  than  5  times  as  great  as  that  of  Mercury  from  the 
sun.  Mars  is  105  million  miles  farther  from  the  sun  than 
Mercury.     What  is  the  distance  of  each  planet  from  the  sun  ? 


20  INTRODUCTION   TO   THE  EQUATION 

38.  The  population  of  New  York  City  (estimate  of  Census 
Bureau,  1904)  was  5  thousand  more  than  11  times  that  of 
Pittsburg.  If  21  times  the  population  of  Pittsburg  is  sub- 
tracted from  twice  that  of  New  York,  the  remainder  is  363 
thousand.     Find  the  population  of  each  city. 

39.  The  standing  army  of  France  (1906)  was  383  thousand 
less  than  twice  that  of  Germany.  If  7  times  the  number  of 
men  in  the  German  army  is  subtracted  from  6  times  the  num- 
ber of  men  in  the  French  army,  the  remainder  is  177  thousand. 
Find  the  number  of  men  in  each  army. 

DIVISION  OF  A  PRODUCT 

22.  The  division  of  the  product  of  several  factors  by  a  given 
number  may  be  performed  in  various  ways  : 

E.g.  (4- 6 -10) --2  =  240-=- 2  =  120. 

Also  (4.6.10)h-2  =  2-6.10  =  120, 

(4.6.10)-2  =  4.3-10  =  120, 
and  (4- 6- 10)- 2  =  4- 6- 5  =  120. 

In  each  case  only  one  factor  is  divided,  and  since  any  factor 
may  be  selected,  we  naturally  choose  one  which  exactly  contains 
the  divisor  if  possible. 

Perform  each  of  the  following  divisions  in  more  than  one 
way  where  possible : 

1.  (5  -8-  3)  -=-2.       4.    (11  •  20  •  16) -- 4.      7.    (10  .  35  .  3)  -5. 

2.  20  abc-h  4.  5.    14  xyz '-*-  7.  8.    14  xyz  -=- x. 

3.  12a5e-i-3.  6.    12  abc  +  c.  9.    (12  •  40  •  13) -s- 8. 

These  examples  illustrate  the  following  principle  : 

23.  Principle  V.  To  divide  the  product  of  several  factors 
by  a  given  number  divide  any  one  of  the  factors  by  that 
number,  leaving  the  other  factors  unchanged. 


DIVISION  OF  A  PRODUCT  21 

Principle  V  is  already  known  in  arithmetic  in  the  process 

called  cancellation. 

2  •  6  •  9 
Thus,  in  the  fraction  — ,  3  may  be  canceled  out  of  either  6 

o 

or  9,  giving  2'6'9  =  2  . 2  .  9  or  2  •  6  . 3. 

O 

Principle  V  is  necessary  in  the  solution  of  problems  like  the 
following : 

Illustrative  Problem.  There  are  three  numbers  whose  sum 
is  26.  The  second  is  8  times  the  first,  and  the  third  is  one-half 
the  second.     Find  each  of  the  numbers. 

Solution.         Let  x  represent  the  first  number, 
then  8  x  represents  the  second  number, 

and  —  represents  the  third  number. 

8a; 
Hence,  x  +  8  x  -\ =  26,  the  sum  of  the  numbers. 

By   Principle  V,     x  +  8  x  +  4  x  =  26,  since  8  x  +  2  =  4  x. 

By  Principle  I,  13  x  =  26. 

Hence,  x  =  2,  the  first  number, 

8  x  =  16,  the  second  number, 

o_       8.9 

and  —  =  — —  =  8,  the  third  number. 

2         2 


EXERCISES  AND  PROBLEMS 

1.  Multiply  2  x  +  3  y  by  5.     Check  for  x  =  7,  y  =  11. 

2.  Multiply  3  a  +  y  by  8.     Check  f or  a  =  2,  y  =  3. 

3.  Divide  3  •  7  •  18  by  6  by  means  of  Principle  V. 

4.  Multiply  7  •  56  •  5  by  6  by  means  of  Principle  III. 

5.  Divide  21  •  36  •  42  by  7,  leaving  the  result  in  two  differ- 
ent forms. 

6.  Multiply  5-13-27  by  3,  leaving  the  result  in  three  dif- 
ferent forms. 


22  INTRODUCTION   TO   THE  EQUATION 

7.  Divide  7  a  •  14  b  •  21  c  by  7  in  three  different  ways. 

8.  Add  5  a,  1-^,  %2,  and  1|^,  using  Principles  V  and  I. 

9.  From  —p^-  subtract        x^,  using  Principles  V  and  II. 

, _     ^         14 a  .  10 a      i  ,       ,6a 

10.  Irom  — —  H — — -  subtract—— 

— ■  o  o 

i ,     -pi-   j  j-v,  ,.  16  a?    20  a;   16  a;    „.  •,  Q 

11.  lind  the  sum  of , , ,  7  a,  and  3  a:. 

8         5         4 

,.    •  ja  n  100  rs   90  rs        ,  25  rs 

12.  Find  the  sum  of  , ,  and  — - — • 

10    '     9  5 

13.  From  25  xy  subtract  — — 2— 

z 

i-i      a  j  a  18  abc  OA ,         -,  30  be 

14.  Add  ,  20  be,  and 

a  '  10 

15.  Add , ,  and • 

s  a  u 

16.  From  i^M  +  17a  subtract  ^^. 

i  M    -c         -(^n       .39  mn      ,  ,       .  25  am 

17.  From  179  m  -\ ■  subtract 

n  a 

18.  From  M(a  +  b) +  U(a  +  b)  subtract  7  (a  +  &). 

19.  Cleveland  had  (estimate  of  Census  Bureau,  1904)  8  times 
as  many  inhabitants  as  Portland,  Maine.  If  twice  the  popu- 
lation of  Portland  is  added  to  ^  that  of  Cleveland,  the  sum  is 
212  thousand.     Find  the  population  of  each  city. 

20.  One  cubic  foot  of  a  certain  kind  of  brick  weighs  as 
much  as  3  cubic  feet  of  cedar.  The  combined  weight  of  ^  of  a 
cubic  foot  of  brick  and  5  cubic  feet  of  cedar  is  187  pounds. 
Find  the  weight  per  cubic  foot  of  each. 


DIVISION  OF  A   SUM  OR  DIFFERENCE  23 

21.  It  is  8  times  as  far  from  Philadelphia  to  Louisville  as 
from  Philadelphia  to  Baltimore.  If  \  the  distance  from  Phila- 
delphia to  Louisville  is  added  to  3  times  that  from  Philadel- 
phia to  Baltimore,  the  sum  is  485  miles.  Find  each  of  the  two 
distances. 

22.  Coinage  silver  weighs  4  times  as  much  per  cubic  inch  as 
feldspar.  The  combined  weight  of  \  a  cubic  inch  of  silver  and 
7  cubic  inches  of  feldspar  is  .846  pound.  What  is  the  weight 
of  a  cubic  inch  of  each  ? 

23.  It  is  3  times  as  far  from  New  York  to  Washington,  D.C., 
as  from  New  York  to  New  Haven,  and  it  is  14  times  as  far 
from  New  York  to  Seattle  as  from  New  York  to  Washington. 
If  \  the  distance  from  New  York  to  Seattle  is  added  to  5  times 
that  from  New  York  to  New  Haven,  the  sum  is  836  miles. 
Find  the  distance  from  New  York  to  each  of  the  other  cities. 

24.  The  population  of  Washington,  D.C.  (estimate  of  Cen- 
sus Bureau,  1904),  was  9  times  that  of  Galveston,  Texas,  and 
the  population  of  Savannah,  Georgia,  was  33  thousand  less 
than  \  that  of  Washington.  The  combined  population  of  Gal- 
veston and  Savannah  was  99  thousand.  Find  the  popidation 
of  each  city. 

25.  A  cubic  foot  of  steel  weighs  17  times  as  much  as  a  cubic 
foot  of  yellow  pine.  The  combined  weight  of  11  cubic  feet  of 
pine  and  3  cubic  feet  of  steel  is  1773.2  pounds.  Find  the 
weight  of  1  cubic  foot  of  each. 

DIVISION  OF  THE  SUM  OR  DIFFERENCE  OF  TWO  NUMBERS 
24.    In  dividing  the  sum  or  difference  of  two  numbers  by  a 
given  number,  when  these  are  represented  by  Arabic  figures, 
the  process  may  be  carried  out  in  two  ways.     Thus, 

(1)  (12  +  8)  --2=20-5-  2  =  10, 

or  (12  +  8)  -  2  =  12  +  2  +  8  +  2  =  6  +  4  =  10. 

(2)  (20  -  12)  +  4  =  8  -  4  =  2, 

or  (20  -  12)  -s-  4  =  20  +  4  -  12  +  4  =  5  -  3  =  2. 


24  INTRODUCTION   TO   THE  EQUATION 

The  same  result  is  obtained  in  each  case  whether  the  num- 
bers in  the  parenthesis  are  first  added  (or  subtracted)  and  the 
result  then  divided  by  the  given  number;  or  the  numbers  in 
parenthesis  are  first  divided  separately  and  then  the  quotients 
added  (or  subtracted). 

If  the  numbers  in  the  dividend  are  represented  by  letters, 
the  division  can  usually  be  carried  out  only  in  the  second 
manner  shown  above. 

E.g.    (r  +  t)+5  =  r  +  5+t  +  5,or,r-±l  =  ^  +  ±. 

o         5     o 

This  is  read :  the  sum  of  r  and  t  divided  by  5  equals  r  divided  by  5 

plus  t  divided  by  5. 

In  this  manner  perform  each  of  the  following  divisions  in 
two  ways  when  possible : 

1.  (16  +  12)  -h  4.  8.  <»  +  $  +  at)  ■*-  a 

2.  (20a- 10  6)  -  5.  7.  (15«  -  3t  +  18t)+3. 

3.  (36  x-  24^/)  -s-6.  8.  (18  a?  +  36  y  -  21 1)  -s-  3. 

4.  (108y-72«)-s-12.  9.  (m  -  n  -  r)  -s-  a. 

5.  (50  x  +  75  x)  -5-  25.  10.  (m  +  n  +  r)  -e-  b. 

These  examples  illustrate  the  following  principle : 

25.  Principle  VI.  To  divide  the  sum,  or  difference  of 
two  numbers  by  a  given  number  divide  each  number  sep- 
arately and  find  the  sum  or  difference  of  the  quotients. 

Principle  VI  is  necessary  in  problems  such  as  the  following : 

Illustrative  Problem.  The  population  of  Iceland  is  40,500  less 
than  10  times  that  of  Greenland.  The  population  of  Green- 
land plus  \  that  of  Iceland  is  27,600.  Find  the  population  of 
each. 


DIVISION  OF  A   SUM  OR  DIFFERENCE  25 

Solution.     Let  x  be  the  number  of  inhabitants  in  Greenland. 

Then  10  x  —  40500  is  the  number  in  Iceland. 

u  ,  10  x  -  40500     ___-. 

Hence  x  + =27600. 

5 

By  Principle  VI,  x  +  2  x  -  8100  =  27600. 

By  Principle  I,  3  x  -  8100  =  27600. 

Adding  8100  to  each  member  and  observing  that  8100  —  8100  =  0, 

3  x  =35700. 
Therefore  x  =  11900,  the  population  of  Greenland, 

and  10  x  —  40500  =  78500,  the  population  of  Iceland. 

Check.  1 1 900  +  ^^  =  27600. 

5 

26.   Principles  V  and  VI  should  be  carefully  contrasted  : 

Thus :      12  * 1S  ' 24  =  4- 18  •  24  =  12  •  6  •  24  =  12  •  18  •  8, 
o 

whi,e  12  +  ^  +  24=4  +  6  +  8. 

That  is,  in  dividing  the  product  of  several  numbers  we  operate 
upon  any  one  of  them  as  found  convenient,  but  in  dividing  the 
sum  of  several  numbers  we  must  operate  upon  each  of  them. 


EXERCISES  AND  PROBLEMS 

1.  Divide  72  +  56  by  8  without  first  adding. 

2.  Divide  144  —  36  by  12  without  first  subtracting. 

3.  Divide  r  +  t  by  5  and  check  the  quotient  when  r  =  15, 
t  =  25;  also  when  r  =  60,  t  =  75. 

4.  Multiply  7  +  9  by  3  without  first  adding  7  and  9. 

5.  Multiply  25  —  8  by  5  without  first  subtracting. 

6.  Find  the  product  of  12  and  a  +  b,  checking  the  result 
when  a  =  5,  b  —  1. 


26  INTRODUCTION   TO   THE  EQUATION 

Perform  the  following  indicated  operations : 

7.  3  (a  +  b  +  c  +  d).     Check  for  a  =  1,  b  =  2,  c  =  3,  d  =  4. 

8.  7(r—  s-\-t  —  x).     Check  for  r  =  t  =  5,  s  =  x  =  4. 

9.  (m  +  n  +  r)  -h  4.      Check  for  ra  =  64,  n  =  32,  »•  =  8. 

10.  (x  +  y  +  z)  -e-  5.       Check  for  a?  =  100,  y  =  50,  and  2  =  25. 

11.  Add  6(o  +  b  +  c)  and  6a  +  126  +  24c. 

o 

12.  From  25  (x  +  y  +  «)  subtract  10°  x  +  50  ?  +  25  z. 

13    Add  6  ft6  +  7  ac  +ad  and  12  h  +  15  c  +  9  d 
a  3 

14.  Perform  the  divisions:  S3x~^V  and  ™t-39s, 

11  13 

15.  Divide  7  ma$  +  3  wa;?/  —  2  ram/  by  y,  and  check  for  ra  =  2, 

n  =  3,  x  =  4,  y  =  5. 

16    Add  21  CTa?  +  49  5y  +  56  a&       ,  24  aa;  +  56  by  +  64  aft 
7  n(  8  ' 

and  check  for  a=b  =  c  =  2,  x  =  y  =  3. 

17.  If  the  sum  of  4  times  a  number  and  32  be  divided  by  2 
the  result  is  30.     Find  the  number. 

18.  The  melting  temperature  of  glass  is  548  degrees  (Centi- 
grade) lower  than  4  times  that  of  zinc.  One-half  the  number 
of  degrees  at  which  glass  melts  plus  7  times  the  number  at 
which  zinc  melts  equals  3434.    Find  the  melting  point  of  each. 

19.  The  melting  temperature  of  silver  is  496  degrees  (Centi- 
grade) lower  than  that  of  nickel.  Five  times  the  number  of 
degrees  at  which  nickel  melts  plus  7  times  the  number  at 
which  silver  melts  equals  13,928.  Find  the  melting  point  of 
each  metal. 


DIVISION  OF  A   SUM  OR  DIFFERENCE  27 

20.  The  population  of  Paris  (1904)  was  1360  thousand  less 
than  twice  that  of  Berlin.  The  sum  of  their  populations  was 
4730  thousand.     Find  the  population  of  each  city. 

21.  A  cubic  foot  of  nickel  weighs  1288  pounds  less  than 
4  cubic  feet  of  tin.  One-half  a  cubic  foot  of  nickel  plus  1  cubic 
foot  of  tin  weighs  724  pounds.  Find  the  weight  per  cubic  foot 
of  each  metal. 

22.  A  cubic  foot  of  gold  weighs  2730  pounds  less  than  6 
cubic  feet  of  silver.  One-third  of  a  cubic  foot  of  gold  together 
with  5  cubic  feet  of  silver  weighs  3675  pounds.  Find  the  weight 
per  cubic  foot  of  each  metal. 

23.  The  population  of  Japan  (1904)  was  106  million  less 
than  12  times  that  of  Manchuria.  If  the  population  of  Man- 
churia be  subtracted  from  \  that  of  Japan,  the  remainder  is  12 
million.     Find  the  population  of  each  country. 

24.  The  population  of  Panama  (1900)  was  1160  thousand 
less  than  3  times  that  of  Nicaragua  (1905).  Three  times  the 
population  of  Panama  plus  twice  that  of  Nicaragua  is  2020 
thousand.     Find  the  population  of  each  country. 

25.  The  population  of  the  Philippines  (1903)  was  400  thou- 
sand less  than  50  times  that  of  Hawaii.  One  twenty-fifth  of 
the  population  of  the  Philippines  plus  the  population  of  Hawaii 
is  464  thousand.     What  was  the  population  of  each  ? 

26.  One  gallon  of  benzine  weighs  45.4  pounds  less  than  8 
gallons  of  alcohol.  The  weight  of  \  a  gallon  of  benzine  and 
2  gallons  of  alcohol  is  16.9  pounds.  Find  the  weight  of  one 
gallon  of  each  liquid. 

27.  During  the  season  1906  the  American  League  baseball 
team  of  Chicago  won  201  games  less  than  6  times  as  many  as 
Boston.  One-third  the  number  of  games  won  by  Chicago  plus 
7  times  the  number  of  games  won  by  Boston  equals  374.  How 
many  games  did  each  team  win  ? 


28  INTRODUCTION   TO   THE  EQUATION 

28.  The  altitude  of  Aconcagua  is  70,800  feet  less  than  6 
times  that  of  Mt.  Blanc.  One-sixth  the  altitude  of  Aconcagua 
added  to  that  of  Mt.  Blanc  equals  19,760  feet.  Find  the  alti- 
tude of  each  mountain. 

29.  The  area  of  Great  Britain  is  30,387  square  miles  less 
than  12  times  that  of  the  Netherlands,  and  the  area  of  Japan 
is  42,065  square  miles  less  than  15  times  that  of  the  Nether- 
lands. One-third  the  area  of  Great  Britain  plus  £  the  area 
of  Japan  is  69,994  square  miles.  Find  the  area  of  each 
country. 

30.  The  diameter  of  the  earth  is  1918  miles  more  than 
twice  that  of  Mercury,  and  the  diameter  of  Venus  is  1700 
miles  more  than  twice  that  of  Mercury.  The  diameter  of  the 
earth  plus  \  that  of  Venus  equals  11,768  miles.  Find  the 
diameter  of  each  planet. 

31.  The  diameter  of  Jupiter  is  500  miles  more  than  20 
times  that  of  Mars,  and  the  diameter  of  Saturn  is  4200  miles 
more  than  16  times  that  of  Mars.  One-tenth  the  diameter  of 
Jupiter  plus  \  that  of  Saturn  is  45,150  miles.  Find  the 
diameter  of  each  planet. 

32.  The  diameter  of  Neptune  is  29,000  miles  less  than 
twice  that  of  Uranus.  One-half  the  diameter  of  Neptune  plus 
4  times  that  of  Uranus  is  145,000  miles.  Find  the  diameter 
of  each  planet. 

From  the  last  three  problems  make  a  table  of  the  diameters 
of  all  the  planets. 

NUMBER   EXPRESSIONS   IN   PARENTHESES 
27.    When  a  number  expression  is  inclosed  in  a  parenthesis, 
it  is  sometimes  possible  to  perform  the  operations  indicated 
within  the  parenthesis  and  thus  to  remove  it. 

E.g.    6  +  (7  +  10  -  9)  =  6  +  8  =  14. 


REMOVAL   OF  PARENTHESES  29 

When,  however,  the  operations  within  the  parenthesis  can- 
not be  carried  out,  the  parenthesis  may  sometimes  be  removed 
by  Principle  IV  or  VI. 

E.g.   5(2  n  +  7  r)  =  10  n  +  35  r,  and  (20  x  -  16  y)  +  4  =  5  x  -  4  y. 

In  these  examples  the  operations  upon  the  parentheses  are 
multiplication  or  division.  The  cases  in  which  the  operations 
on  the  parentheses  are  addition  or  subtraction  are  considered 
in  the  following  examples : 

(1)  5  +  (3  +  4)  =  5  +  7  =  12, 

also  5  +  (3  +  4)  =  5  +  3  +  4  =  8  +  4  =  12. 

(2)  10  +  (7-3)  =  10  +  4  =  14, 

also  10  +  (7-3)=10  +  7-3  =  17-3  =  14. 

(3)  20- (7 +2)  =20-9  =  11, 

also  20  -  (7  +  2)  =  20  -  7  -  2  =  13  -  2  =  11. 

(4)  20-(7-2)=20-5  =  15. 

also  20-(7-2)=20-7+2  =  13  +  2  =  15. 

From  (1)  it  appears  that  the  expression,  3  +  4,  may  be  added 
to  5  by  first  adding  3  and  then  adding  4. 

From  (2)  it  is  seen  that  the  expression,  7  —  3,  may  be  added 
to  10  by  first  adding  7  and  then  subtracting  3. 

From  (3)  it  is  seen  that  the  expression,  7  +  2,  may  be  sub- 
tracted from  20  by  first  subtracting  7  and  then  subtracting  2. 

From  (4)  it  is  evident  that  the  expression,  7  —  2,  may 
be  subtracted  from  20  by  first  subtracting  7  and  then  adding  2, 
since  7  is  2  more  than  the  number  which  was  to  be  subtracted. 

In  like  manner  perform  each  of  the  following  operations  in 
two  ways : 

1.  18 +  (5  +  2  +  3).  4.  40 -(6 +  7+3).       7.  25n+(22w-3rc). 

2.  28 +  (7 -3 +  4).  5.  55 -(16 -7).  8.  18*-(6a;-2aj). 

3.  32 -(5 +  3).  6.  101 +  (73 -22).       9.  16r-(7r  +  2r). 


30  INTRODUCTION  TO   THE  EQUATION 

These  examples  illustrate  the  following  principle : 
28.  Principle  VII.  A  number  expression  consisting  of 
two  or  more  numbers  connected  by  the  signs  +  or  —  may 
be  added  to  anotiier  given  number  by  adding  each  number 
with  the  plus  sign  and  subtracting  each  number  with  the 
minus  sign.  Such  a  number  expression  may  be  sub- 
tracted from  a  given  number  by  subtracting  each  number 
with  the  plus  sign  and  adding  each  number  with  the 
minus  sign. 

Principles  IV,  VI,  and  VII  are  useful  in  removing  parenthe- 
ses from  number  expressions  when  the  values  of  the  numbers 
involved  are  not  known. 

Instead  of  a  parenthesis,  a  bracket  [     ],  or  a  brace  {    },  may  be  used. 
Thus,  2  (6  +  4)  =  2  [6  +  4]  =  2  {6  +  4}. 

In  expressions  involving  parentheses  the  operations  within 
the  parentheses  should  be  performed  first  if  possible.  Then 
perform  the  indicated  multiplications  and  divisions,  and  finally 
the  remaining  additions  and  subtractions. 

E.g.     Given  7  +  15  -  (9  -  4)  -  4  [7  +  11]  +  (29  -  20). 

Performing  operations  within  the  parentheses,  this  reduces  to : 
7  +  15-5-4-18^9. 

Performing  multiplications  and  divisions,  we  have  7  +  3  —  8. 

Performing  the  remaining  additions  and  subtractions,  the  given 
expression  reduces  to  2. 

EXERCISES 

In  performing  the  following  indicated  operations,  state  in 
each  case  what  principles  are  used. 

1.8(7+4  +  8  +  2).  8.  6(7-3)+2. 

3.   3(2a-4)-4(a-6).       4.  14  +  4  (8  +  4)  -=-  (32-20). 

The    minus    sign   preceding   4  (a  —  6)  indicates  that  this  whole 

product  is  to  be  subtracted.     Hence,  using  Principle  IV,  we  have 

6a—  12  —  (4a  —24).      Then,    using    Principle    VII,   this    becomes 

6a-12-4a  +  24  =  2a  +  12. 


REMOVAL    OF  PARENTHESES  31 

5.  8-3(4-2)+6  +  3[6  +  l]. 

6.  16  -5- [3  +  1]-  [12 -3j  -4-3. 

7.  5  (17  -  5)  +  18  -=-  (8  -  2)  -  (21  +  14)  -=-  7. 

8.  7(a  +  6)+6(a  +  6). 

9.  5(a+6)+  3  (a  +  b  +  c). 

10.   16(r  +  *)  +  H  {r4  0-  11.  15s-3(r  +  s). 

12.    15  (r  +  s)  -3(r-s).  13.   20-(cc  +  w), 

14.   50x-25(x-y).  15.    12  m  -  6(*  -  2m). 

16.  ll<  +  5(2*  -1)  -3(2  +  2).     Check  fori  =  2. 

17.  (12* -6m)  -r-2  +  3«  — 2m.     Check  for  t  =  1,  u  =  1. 

18.  3(5a-7y)  +  (21a-28M)  -e- 7.    Check  for  a;  =  2,  m  =  1. 

19.  8(r  —  s)  +  5(2r  +  s)  —  3 (r  +  s).   Check  for  r  =  1,  8  =  1. 

20.  10(a;  +  y)  -r-  5  +  6 (a?  -  y)  -*-  3.     Check  f or  cc  =  2,  m  =  1. 

21.  5(h+k)  +  2(h+k)+3(h  +  7c). 

Use  Principle  I,  then  IV ;  also  IV,  then  I. 

22.  5(7x-4:y)+9(9x-5y). 

23.  8(r-s)-2(r-s).  24.   9(3p- q)-S(3p-q). 
Use  Principle  II,  then  IV ;  also  reverse  this  order. 

25.    (16  x  -12  x)  -f-  4.  26.    (18  a6  — 12  a6a?) -s- a. 

Use  first  Principle  VI,  then  II ;  also  II,  and  then  V. 

27.   9(6 +  4) -3  (6 -4).  28.    (5  axy  -  3  ex) -=-  x. 

29.  9  (6  abc  +  4:  xyz)  —  3(6  abc  —  4  an/z). 

30.  8(5a;-M  +  2z)-ll(3a;  +  2M-7z). 

31.  3[2(a  +  6  +  c)-3(a-6  +  c)]. 

First  remove  the  parentheses,  then  the  bracket. 

32.  7\Sx-(2y-3x)+(2x-Ay)'i. 


32  INTRODUCTION   TO   THE  EQUATION 

PROBLEMS 

1.  There  are  three  numbers  whose  sum  is  80.  The  second 
is  3  times  the  first,  and  the  third  twice  the  second.  What  are 
the  numbers  ? 

2.  A  man  has  three  buildings  whose  total  value  is  $46,800. 
The  second  building  cost  $  800  less  than  the  first,  and  the  third 
cost  twice  as  much  as  the  second.  What  is  the  cost  of  each 
building  ? 

3.  The  population  of  Connecticut  (Census  of  1900)  was  50 
thousand  more  than  twice  that  of  Rhode  Island,  and  the  popu- 
lation of  Massachusetts-  was  81  thousand  more  than  3  times 
that  of  Connecticut.  The  population  of  Massachusetts  minus 
that  of  Connecticut  was  1897  thousand.  Find  the  population 
of  each  state. 

4.  The  population  of  Indiana  (Census  of  1900)  exceeded  that 
of  Iowa  by  284  thousand,  and  the  population  of  Illinois  was 
210  thousand  less  than  twice  that  of  Indiana.  The  population 
of  Illinois  minus  that  of  Indiana  was  2306  thousand.  Find 
the  population  of  each  state. 

5.  The  melting  point  of  copper  is  250  degrees  (Centigrade) 
lower  than  4  times  that  of  lead.  Ten  times  the  number  of 
degrees  at  which  lead  melts  minus  twice  the  number  at  which 
copper  melts  equals  1152.  What  is  the  melting  point  of  each 
metal  ? 

6.  The  melting  point  of  iron  is  450  degrees  higher  than  5 
times  that  of  tin.  Three  times  the  number  of  degrees  at 
which  iron  melts  plus  7  times  the  number  at  which  tin  melts 
equals  6410.     Find  the  melting  point  of  each  metal. 

7.  In  1904  the  gold  product  of  Africa  was  11  million  dol- 
lars more  than  3  times  that  of  Russia,  and  the  gold  product  of 
Australia  was  84  million  less  than  twice  that  of  Africa. 
Russia  and  Australia  together  produced  113  million.  How 
much  did  each  country  produce  ? 


REMOVAL    OF    PARENTHESES  33 

8.  In  1904  the  value  of  the  silver  produced  in  the  United 
States  was  5  million  dollars  more  than  14  times  as  much  as 
that  of  Canada,  and  the  product  of  Mexico  was  1  million  less 
than  16  times  that  of  Canada.  Twice  the  product  of  the 
United  States  minus  that  of  Mexico  equals  71  million.  How 
much  did  each  country  produce  ? 

9.  The  Nile  is  100  miles  more  than  twice  as  long  as  the 
Danube.  Ten  times  the  length  of  the  Danube  plus  4  times  the 
length  of  the  Nile  equals  3400  miles.    How  long  is  each  river  ? 

10.  The  number  of  first-class  battleships  in  the  United  States 
navy  (1906)  was  22  less  than  5  times  the  number  of  protected 
cruisers.  Twelve  times  the  number  of  cruisers  less  twice  the 
number  of  battleships  equals  60.     Find  the  number  of  each. 

11.  The  number  of  torpedo  boats  in  the  United  States  navy 
(1906)  was  77  less  than  7  times  .as  great  as  the  number  of 
torpedo  boat  destroyers.  Three  times  the  number  of  torpedo 
boats  minus  4  times  the  number  of  destroyers  equals  41.  Find 
the  number  of  each. 

12.  The  weight  of  a  cubic  foot  of  spruce  is  16  pounds  more 
than  that  of  a  cubic  foot  of  cork,  and  the  weight  of  a  cubic 
foot  of  dry  live  oak  is  5  pounds  more  than  twice  that  of  spruce. 
One  cubic  foot  of  oak  weighs  36  pounds  more  than  one  cubic 
foot  of  spruce.     Find  the  weight  of  a  cubic  foot  of  each. 

13.  In  1904  Canada  produced  3  million  dollars  more  of 
gold  than  Mexico,  and  the  United  States  produced  17  million 
more  than  4  times  as  much  as  Canada.  The  combined  product 
of  Mexico  and  the  United  States  was  94  million.  How  much 
did  each  country  produce  ? 

14.  The  value  of  the  copper  produced  in  the  United  States 
in  1904  was  5  million  dollars  more  than  the  value  of  the  crude 
petroleum,  and  the  value  of  the  bituminous  coal  was  94  million 
more  than  twice  the  value  of  the  copper.  Nine  times  the  value 
of  the  copper  minus  twice  the  value  of  the  coal  equals  342 
million.     Find  the  value  of  each. 


34  INTRODUCTION  TO   THE  EQUATION 

15.  The  pressure  developed  in  the  chamber  of  a  modern 
twelve-inch  gun  is  21  thousand  pounds  per  square  inch  more 
than  that  developed  in  an  ordinary  hunting  rifle.  The 
maximum  pressure  which  gunpowder  can  develop  is  18 
thousand  pounds  per  square  inch  less  than  3  times  as  great 
as  that  developed  in  the  twelve-inch  gun.  The  sum  of 
the  three  pressures  is  151  thousand  pounds.  Find  each 
pressure. 

16.  The  standing  army  of  Australia  (1906)  was  14  thousand 
more  than  that  of  Canada,  and  the  standing  army  of  Great 
Britain  was  47  thousand  more  than  4  times  that  of  Australia. 
Great  Britain's  army  contained  287  thousand  men.  How  many 
men  were  there  in  the  armies  of  Canada  and  Australia  ? 


IDENTITIES  AND   EQUATIONS 

29.  In  the  preceding  pages  equations  have  been  freely 
used,  and  certain  simple  methods  have  been  employed  in 
solving  problems  by  means  of  them.  We  now  proceed  to  a 
more  detailed  study  of  these  methods  and  of  the  equation 
itself. 

30.  Equalities  in  which  letters  are  used  as  number  symbols 
are  of  two  kinds  as  shown  by  the  following  examples: 

(1)  3n  +  4n  =  7»isa  true  statement  no  matter  what  number  is 
represented  by  n.  3  (5  x  +  6y)  =  15  x  +  18  ?/  holds  for  all  values 
which  may  be  assigned  to  x  and  y. 

(2)  to  +  2  (w  +  5)  =  58  is  a  true  statement  if,  and  only  if,  w  =  16.  If 
w  is  replaced  by  any  number  less  than  16,  the  number  expression  on 
the  left  is  less  than  58,  and  if  w  is  replaced  by  any  number  greater 
than  16,  the  result  is  greater  than  58. 

31.  The  equality  w  -f  2  (w  -f  5)  =  58  is  said  to  be  satisfied  by 
w=16,  because  this  value  of  w  reduces  both  members  to  the 
same  number,  58. 


IDENTITIES  AND  EQUATIONS  35 

32.  Definition.  An  equality  which  is  satisfied,  no  matter 
what  numbers  are  substituted  for  one  or  more  of  its  letters, 
is  called  an  identity  with  respect  to  those  letters. 

E.g.  3n  +  4n  =  7nisan  identity  with  respect  to  n. 

a  (b  +  c)  =  ab  +  ac  is  an  identity  with  respect  to  a,  b,  and  c. 

When  it  is  especially  desired  to  distinguish  an  equality  as  an 
identity,  the  equality  sign  is  written  =. 

Each  of  the  Principles  I  to  VII  may  be  thus  stated  in  symbols  as 
identities.    Thus, 

I,  4n  +  6n  =  10n;         II,  12  n  -  5n=7n; 
III,  5  (4  ab)=20  ab ;         IV,  5  (a  ±  b)=5  a±5b; 
V,  30a&^-6  =  5aZ>;         VI,  (16  a  ±  206)  + 4=4  a  ±  56; 
VII,  x  +  (a  -  b)  -  (c  -  d)=x  +  a-b-c  +  d. 

33.  Definition.  An  equality  which  is  satisfied  only  when 
certain  particular  values  are  given  to  one  or  more  of  its  letters 
is  called  a  conditional  equality  with  respect  to  those  letters. 

E.g.  3  x  +  5  =  35  is  an  equality  only  on  the  condition  that  x  =  10. 
x  +  y  =  10  is  an  equality  for  certain  pairs  of  letters  like  1  and  9, 2  and  8, 
3  and  7,  5  and  5,  but  certainly  not  for  all  values  of  x  and  y ;  for  instance, 
not  for  3  and  8. 

34.  A  conditional  equality  is  called  an  equation,  and  a  letter 
whose  particular  value  is  sought  to  satisfy  an  equation  is  called 
an  unknown  number  in  that  equation,  or  simply  an  unknown. 

The  equations  at  present  considered  contain  only  one  un- 
known. 

35.  To  solve  an  equation  in  one  unknown  is  to  find  the  value 
or  values  of  the  unknown  which  satisfy  it. 

The  following  example  exhibits  the  process  of  solving  an 
equation  which  contains  one  unknown  and  which  is  satisfied 
by  one  value  only  of  the  unknown : 


% 


36  INTRODUCTION   TO   THE  EQUATION 

Given  w  +  2  (w  +  5)  =  58.  (1) 

By  Principle  III,  to  +  2  to  +  10  =  58.  (2) 

By  Principle  I,  3  to  +  10    =  58.  (3) 

Subtracting  10  from  both  members,      3  to  =  48.  (4) 

Dividing  both  sides  by  3,  w  =  16.  (5) 

Check.    Putting  w  =  16  in  (1),     16  +  2  (16  +  5)  =  16  +  2  .  21  =  58. 

Explanation.    Any  value  of  to  which  satisfies  (1)  also  satisfies  (2) 

and  (3),  since  the  expressions  in  the  left  members  of  (1),  (2),  and  (3) 

represent  the  same  number  expressed  in  different  forms.     Any  value 

of  to  which  satisfies  (3)  also  satisfies  (4),  for  if  10  more  than  3  to  is 

58,  then  3  to  must  be  48.     Any  value  of  to  which  satisfies  (4)  also 

satisfies  (5),  for  if  3  to  is  48,  then  to  is  |  of  48. 

By  similar  considerations  the  value  of  to  which  satisfies  (5)  may  be 
shown  to  satisfy  (4),  (3),  (2),  and  (1).  Hence  to  =  16  is  the  solution 
of  the  given  equation. 

The  above  solution  illustrates  the  following  principle : 
36.   Principle  VIII.    An  equation  may  be  changed  into 
another  equation  such  that  any  value  of  the  unknown 
which  satisfies  one  also  satisfies  the  other,  by  means  of  any 
of  the  following  operations : 

(1)  Adding  the  same  number  to  both  members. 

(2)  Subtracting  the  same  number  from  both  members. 

(3)  Multiplying  both  members  by  the  same  number. 

(4)  Dividing  both  members  by  the  same  number. 

(5)  Changing  the  form  of  either  member  in  any  way 
which  leaves  its  value  unaltered. 

The  operations  under  Principle  VIII  are  hereafter  referred 
to  in  detail  by  means  of  the  initial  letters,  A  for  addition,  S  for 
subtraction,  M  for  multiplication,  D  for  division,  and  F  for 
form  changes. 

Note. — It  is  not  permissible  to  multiply  or  divide  the  members  of 
an  equation  by  any  expression  which  is  equal  to  zero.  See  Advanced 
Course. 


SOLUTION  OF  EQUATION H  37 

37.  The  operations  involved  in  Principles  I  to  VII  are 
all  form  changes  which  leave  the  value  of  the  expression 
unaltered. 

E.g.  4  (vi  +  n)  by  Principle  IV  has  the  same  value  as  4  m  +  4  n. 

There  are  other  form  changes  which  are  already  familiar  in 
arithmetic. 

E.g.  2  +  4  =  4  +  2,  or  in  general  a  +  b  =  b  +  a.  Likewise  2-7  =  7-2, 
or  in  general  ab  =  ba.  That  is,  the  order  in  which  numbers  are  added 
or  multiplied  is  immaterial. 

Again,  3  +  4  +  6  =  (3  +  4)  +  6  =  3  +  (4  +  6),  or  in  general  a  +  b  +  c 
=  (a  +  6)  +  c  =  a  +  (6  +  c),  and  likewise  a -b •  c  =  (a  •  b)  •  c  =  a •  (6 •  c)  ; 
i.e.  numbers  to  be  added  or  multiplied  may  be  grouped  in  any  manner 
desired.     Still  other  form  changes  will  be  learned  later. 

38.  Principle  VIII  is  further  illustrated  as  follows : 

On  the  scale  pans  of  a  common  balance  are  placed  objects  of 
uniform  weight,  say  tenpenny  nails.  The  scales  balance  only 
when  the  weights  are  the  same  in  both  pans ;  that  is,  when  the 
number  of  nails  is  the  same. 

If  now  the  scales  are  in  balance,  they  will  remain  so  under 
two  kinds  of  changes  in  the  weights : 

(a)  When  the  number  of  nails  in  the  two  pans  is  in- 
creased or  diminished  by  the  same  number ;  correspond- 
ing to  the  operations  A,  S,  M,  D,  on  the  members  of  an 
equation. 

(6)  When  the  number  in  each  pan  is  left  unaltered  but 
the  nails  are  rearranged  in  groups  or  piles  in  any  manner ; 
corresponding  to  the  operations  F  on  the  members  of  an 
equation. 

The  equation,  then,  is  like  a  balance,  and  its  members  are 
to  be  operated  upon  only  in  such  ways  as  to  preserve  the 
balance. 


38  INTRODUCTION   TO   THE  EQUATION 

DIRECTIONS  FOR  WRITTEN   WORK 

39.  The  solution  of  an  equation  consists  in  deducing,  by 
means  of  Principles  I  to  VIII,  another  equation  whose  first 
member  contains  the  unknown  only  and  whose  second  member 
does  not  contain  the  unknown.  The  successive  steps  should 
be  written  down  as  in  the  following  example  : 

Solve  6  (4  n  -  3)  +  25 (n  +  1)  =  50  +  31  ft  +  2 (3  -  ft)-  9.  (1) 

By  F,  using  III  and  IV,  we  obtain  from  (1) 

24  n  -18  +  25  n  +  25  =  50  +  31n  +  6-2  n-  9.  (2) 

By  F,  I  and  II,  we  obtain  from  (2) 

49  »  +  7  =  29  n  +  47.  (3) 

Subtracting  7  and  29  n  from  each  member  of  (3)  and  using  Prin- 
ciple II,  we  have  20n  =  40.  (4) 

Dividing  each  member  of  (4)  by  20, 

n=2.  (5) 

Check.    Substitute  n  =  2  in  Equation  (1). 

For  convenience  this  work  can  be  abbreviated  as  follows: 

6(4n  -  3)  +  25  (n  +  1)  =  50  +  31  n  +  2(3-  n)  -  9.  (1) 

By  F,  III,  IV,  24  n  -  18  +  25  n  +  25  =  50  +  31  n  +  6  -  2  n  -  9.  (2) 
By  F,  I,  II,  49n  +  7  =  29n  +  47.  (3) 

By  S 1 7,  29  n,  20n  =  40.  (4) 

ByD|20,  n  =  2.  (5) 

S  |  7,  29  n  means  that  7  and  29  n  are  each  to  be  subtracted  from 
both  members  of  the  preceding  equation.  D 1 20  means  that  the 
members  of  the  preceding  equation  are  to  be  divided  by  20. 

Similarly,  in  case  we  wish  to  indicate  that  6  is  to  be  added  to  each 
member  of  an  equation,  we  would  write  A  I  6,  and  if  each  member  is 
to  be  multiplied  by  8,  we  would  write  M\  8.  It  is  important  that  the 
nature  of  each  step  be  recorded  in  some  such  manner. 


+  4  n  -  6  =  20  +  5n  -  5  +  3  n. 

(2) 

21  n  +  2  =  15  +  8  n. 

(3) 

21n-8n  =  15-2. 

(4) 

13  n  =  13. 

(5) 

n  =  1. 

(6) 

SOLUTION  OF  EQUATIONS  39 

(2)    Solve  17 n  +  4(2  +  ft)  -  6  =  5(4  +  n)  -  5  +  3  w.       (1) 

By  F,  IV,  17  n 

By  F,  I, 
By  S  |  2,  8  n, 
By  F,  II, 
By  Z>  1 13, 

Check.     Substitute  n  =  1  in  equation  (1); 
then  17  +  12  -  6  =  25  -  5  +  3, 

or  23  =  23. 

An  equation  may  be  translated  into  a  problem.  For  example, 
the  equation  21a;  +  2  =  15  +  8ic  may  be  interpreted  as  follows : 
Find  a  number  such  that  21  times  the  number  plus  2  is  15 
greater  than  8  times  the  number. 

EXERCISES  AND  PROBLEMS 

Solve  the  following  equations,  putting  the  work  in  a  form 
similar  to  the  above  and  checking  each  result.  Translate  the 
first  twenty  into  problems. 

1.  13a-40-tt  =  8. 

2.  3z  +  9  +  2a;  +  6  =  18  +  4:C. 

3.  5  x  +  3  —  x  =  x-^-18. 

4.  13y  +  12  +  5y=32+8y. 

5.  4m  +  6ra  +  4  =  9m  +  6. 

6.  7m-18  +  3m  =  12  +  2m  +  2. 

7.  3y-4;  +  2y-6  =  y  +  7  +  y  +  3  +  10. 

8.  5«  +  3  +  2a;  +  3  =  2a;  +  5  +  3a;  +  3  +  a;. 

9.  2a;  +  4a;4-9  +  a;  +  6  =  20  +  3a;  +  5  +  2£c. 

10.   18  +  6m  +  30  +  6m  =  4m  +  8+12  +  3m  +  3  +  m  +  29. 


40  INTRODUCTION   TO   THE  EQUATION 

11.  y  +  72  +  45y  =  106  +  12y. 

12.  42  a -56  =  20  a: +  10. 

13.  6x  +  8-3x  +  4:  +  5x  =  7x  +  32  +  x-20. 

14.  32x-4  +  7x  =  B8  +  3x  +  5x. 

15.  12m-3-3ra  =  32  +  2m. 

16.  15ra  +  3-2m  +  7  =  3m  +  60. 

17.  a  +  7+3a  =  2a  +  45. 

18.  5  6-30  +  66  =  36  +  90. 

19.  3c  +  18+14c  =  6c  +  51. 

20.  17x-  +  4  +  3x  =  7a;  +  30. 

21.  7(m  +  6)  +  10  m  =  42  -  8(2  m  +  2)  + 181. 

22.  20  -  3(x  -  4)  +  2  x  =  2  x  + 17. 

23.  (8x  +  4)-=-2  +  7a;  =  4z  +  9. 

24.  6  a  +  4(4  x  +  2)  =  85  -  3(2  x  +  7). 

25.  8  +  7(6  +  6  n)  +  2  w  =  2(4  n  +  5)  +  18  n  +  49. 

26.  5(9a:  +  3)  +  6a:  =  24a:-4(3a;  +  2)  +  36. 

27    7(12  «;  +8)  +  13  +  5  a;  -  6  =  47.     Apply  V  and  VI. 
28.    12(5  +  4o;)_5(6  +  4a;)+5-()==a;  +  18j 

29    15  ,  21(3  +  *)     2(6  +  18^)^3(9^+12)     2S 
7  3  3         "*" 

__     ll(5z  +  25),  3(6^-2)      7(4a;  +  8),12a;+36jW 
30. ^  +  — —  -J  = r-^  + _ +35. 


SOLUTION   OF  EQUATIONS  41 


EQUATIONS   INVOLVING   FRACTIONS 

40.  Solve  w  +  ^  +  ^  =  88.  (1) 

First  Solution.     The  coefficients  of  n  are  1,  f,  and  J. 
Applying  Principle  I,  If  n  =  88.  (2) 

By  D  I  If ,  n  =  88  +  If  =  48.  (3) 

Second  Solution.   Multiply  both  members  of  (1)  by  6. 

That  is,  by  1/ 1  6,  6n+^  +  ^=  528.  (2) 

By  F,  V,  6n+3n+2n=  528.  (3) 

By  F,  I,  11  n  =  528.  (4) 

By  D  |  11,  n  =  48,  as  before.  (5) 

The  object  is  to  multiply  both  members  of  the  equation  by  such  a 
number  as  will  cancel_each  denominator.  Hence  the  multiplier  must 
contain  each  denominator  as  a  factor. 

Evidently  12  or  18  might  have  been  chosen  for  this  purpose,  but  not 
8  or  10.  6  is  the  smallest  number  which  will  cancel  both  2  and  3,  and 
hence  this  was  chosen  as  the  multiplier. 

41.  The  process  explained  in  the  second  solution  above  is 
called  clearing  of  fractions. 

As  another  illustration  solve  the  equation 

Here  the  smallest  multiplier  available  is  36. 

Hence  by  M|36,    §^L*  +  ?<L£  +  Hl&£  -  89  •  65i  (2) 

By  V,  each  denominator  is  cancelled  by  the  factor  36, 

9  •  3  x  +  18  x  +  4  •  5  x  =  36  •  65.  (3) 

By  III,                        27  x  +  18  x  +  20  x  =  36  •  65.  (4) 

By  I,                                                   65  x  =  36 .  65.  (5) 

ByD,V,  x  =  §6^65  =  36<  (6) 

o5 
Check.   Substitute  x  =  36  in  (1). 


42  INTRODUCTION   TO   THE  EQUATION 


EXERCISES  AND  PROBLEMS 

Solve  the  following  equations,  indicating  the  principles 
used  at  each  step.  Check  each  solution  by  substituting  in 
the  original  equation  the  value  obtained.  Translate  each 
into  a  problem. 

For  instance,  from  equation  2:  Find  a  number  such  that  when 
increased  by  its  half,  its  third,  and  its  fourth,  the  sum  is  25. 

1.    V-  =  5. 
2^3 

3.    ^  +  4n-^=^  +  26. 

4  3         2 

2     3     4     10 

lb  5      10 

6.  4n  +  ^  =  3ft  +  2 

7  2 

7.  12   |   4(9a;  +  6)      2(3  +  lla;)_5(4a;  +  4)  |   11 

o  o  o 

8.  l3&  +  7-^±^  =  ?A±9  +  38. 

7  5 

5  a  -f  7  .  2a  +  4  =  3a  +  9      ^    . 

2  3  4 

10     17a~ 5     10a  +  2  =  5a  +  7     g 

3  4       "        2 

11.    ^±3  +  5(2a;-fl0)  =  2a;  +  24 
—  5 


SOLUTION  OF  EQUATIONS  43 

12.  The  sum  of  two  numbers  is  12,  and  the  first  number  is 
\  as  great  as  the  second.     What  are  the  numbers  ? 

13.  The  smaller  of  two  numbers  is  f  of  the  larger.  If  their 
sum  is  66,  what  are  the  numbers  ? 

14.  Find  two  consecutive  integers  such  that  4  times  the 
first  minus  3  times  the  second  equals  9. 

15.  Find  three  consecutive  integers  such  that  \  of  the  first 
plus  the  second  minus  ^  the  third  equals  5. 

16.  Find  three  consecutive  integers  such  that  3  times 
the  first  plus  9  times  the  second  minus  4  times  the  third 
equals  73. 

17.  There  are  three  numbers  such  that  the  second  is  4  more 
than  9  times  the  first,  and  the  third  is  2  more  than  6  times 
the  first.  If  -|-  of  the  third  is  subtracted  from  \  of  the  second, 
the  remainder  is  3.     Find  the  numbers. 

18.  There  are  three  numbers  such  that  the  second  is  2  more 
than  9  times  the  first  and  the  third  is  7  more  than  11  times  the 
first.  The  remainder  when  4  times  the  third  is  subtracted  from 
13  times  the  second  is  144.     Find  the  numbers. 

19.  The  population  of  Philadelphia  (estimate  of  Census 
Bureau,  1904)  was  508  thousand  less  than  20  times  that  of 
Dayton,  Ohio.  The  population  of  St.  Louis  was  245  thousand 
more  than  4  times  that  of  Dayton.  One-half  the  population  of 
Philadelphia  plus  ^  that  of  St.  Louis  was  851  thousand.  Find 
the  population  of  each  city. 

20.  The  shortest  railway  route  from  Boston  to  Chicago 
is  166  miles  more  than  4  times  that  from  Boston  to  New 
York ;  and  the  shortest  route  from  Boston  to  Atlanta  is 
196  miles  less  than  6  times  that  from  Boston  to  New  York. 
The  distance  from  Boston  to  Chicago  is  481  miles  more  than 
\  the  distance  from  Boston  to  Atlanta.  Find  each  of  the  three 
distances. 


44  INTRODUCTION  TO   THE  EQUATION 

Solution.  Let  d  —  distance  in  miles  from  Boston  to  New  York  ; 
then  4  d  +  166  =  distance  in  miles  from  Boston  to  Chicago, 

and  6  d  —  196  =  distance  in  miles  from  Boston  to  Atlanta, 

and  4rf  +  166  =  6J~196  +  481. 

By  VI,     4  d  +  166  =  3  d  -  98  +  481. 

By  S,  II,  d  =  481  -  98  -  166  =  217  =  distance  from  Boston 

to  New  York. 

4  d  +  166  =  1034  =  distance  from  Boston  to  Chicago, 

and  6  d  —  196  =  1106  =  distance  from  Boston  to  Atlanta. 

21.  The  railway  mileage  in  the  United  States  in  1900  was 
8  thousand  greater  than  twice  that  of  1880,  and  in  1904  it  was 
66  thousand  less  than  3  times  that  of  1880.  One-half  the  mile- 
age of  1900  plus  \  that  of  1904  equals  168  thousand.  Find 
the  railway  mileage  in  1880,  1900,  and  1904. 

22.  The  number  of  newspapers  in  the  United  States  in  1880 
was  381  less  than  4  times  that  in  1850.  The  number  in  1905 
was  700  more  than  twice  that  of  1880  and  7990  more  than 
6  times  that  of  1850.  Find  the  number  of  newspapers  in  1850, 
1880,  and  1905. 

23.  The  number  of  telephones  in  the  United  States  in  1900 
was  40  thousand  less  than  30  times  as  great  as  the  number  in 
1880.  Half  the  number  in  1905  was  660  thousand  more  than 
that  in  1900  and  80  thousand  more  than  40  times  that  in  1880. 
Find  the  number  of  telephones  in  use  in  1880,  1900,  and  1905. 

24.  The  displacement  of  the  battleship  Kearsarge  is  1232 
tons  greater  than  that  of  the  Oregon,  and  the  displacement 
of  the  Connecticut  is  4480  tons  greater  than  that  of  the 
Kearsarge.  One-third  the  displacement  of  the  Kearsarge 
minus  \  that  of  the  Connecticut  equals  640.  Find  the  dis- 
placement of  each  ship. 

25.  The  distance  of  the  fixed  star  Vega  from  the  sun  is  11.7 
light-years  greater  than  the  distance  from  the  sun  to  Sirius. 
The  distance  of  Regulus  is  8  light-years  less  than  twice  that  of 


SOLUTION  OF  EQUATIONS  45 

Vega.    What  is  the  distance  of  each  star  from  the  sun  if  \  the 
distance  of  Vega  minus  \  that  of  Regulus  is  2  light-years  ? 

Light  travels  at  the  rate  of  186,000  miles  in  one  second.  A  light- 
year  is  the  distance  traveled  by  light  in  one  year. 

26.  The  distance  from  the  sun  to  the  fixed  star  61  Cygni  is 
4  light-years  more  than  that  from  the  sun  to  Alpha  Centauri 
(the  star  nearest  the  sun),  and  the  distance  from  the  sun  to 
Polaris  (the  pole  star)  is  6  times  that  of  61  Cygni.  One- 
eighth  the  distance  of  Polaris  minus  \  the  sum  of  the  dis- 
tances of  the  other  two  equals  2  light-years.  How  many 
light-years  is  each  star  from  the  sun? 

27.  In  1905  the  wheat  crop  of  Russia  was  47  million  bushels 
less  than  4  times  that  of  Argentina,  and  the  crop  of  the  United 
States  was  77  million  bushels  less  than  5  times  that  of  Argentina. 
The  crop  of  the  United  States  exceeded  that  of  Russia  by  124 
million  bushels.     What  was  the  wheat  crop  of  each  country  ? 

28.  The  total  imports  of  the  United  States  in  1900  were  60 
million  dollars  less  than  10  times  the  imports  in  1800.  The 
imports  in  1906  were  473  million  less  than  twice  the  imports  of 
1900,  and  135  million  more  than  12  times  the  imports  of  1800. 
Find  the  imports  of  the  United  States  in  1800,  1900,  and  1906. 

29.  The  total  exports  of  the  United  States  in  1900  were  26 
million  dollars  less  than  20  times  the  exports  in  1800.  The 
exports  in  1906  exceeded  those  of  1900  by  350  million  and  were 
31  million  less  than  25  times  those  of  1800.  Pind  the  exports 
of  1800,  1900,  and  1906. 

30.  The  gross  tonnage  of  the  German  merchant  marine  was 
(1906)  548  thousand  more  than  that  of  the  United  States,  and 
the  tonnage  of  the  British  marine  was  2662  thousand  greater 
than  4  times  that  of  the  German  marine.  The  British  marine 
was  greater  by  930  thousand  tons  than  twice  that  of  Germany 
and  3  times  that  of  the  United  States  combined.  Find  the 
tonnage  of  each  marine. 


46  INTRODUCTION  TO   THE  EQUATION 

31.  The  population  of  San  Francisco  (estimate  of  Census 
Bureau,  1904)  was  289  thousand  more  than  that  of  Springfield, 
Massachusetts,  and  the  population  of  Chicago  was  132  thousand 
more  than  5  times  that  of  San  Francisco.  The  population  of 
Springfield  minus  -£-$  that  of  Chicago  was  2  thousand.  Find 
the  population  of  each  city. 

32.  The  per  capita  tax  of  New  York  City  (1905)  was  $  .49 
less  than  twice  that  of  Chicago  and  the  per  capita  tax  of  Boston 
was  $  3.56  less  than  3  times  that  of  Chicago.  The  combined 
per  capita  tax  of  Boston  and  New  York  was  $  52.15.  What 
was  the  per  capita  tax  of  each  city  ? 

33.  The  number  of  pupils  attending  public  high  schools  in 
Chicago  (1905)  was  1237  more  than  twice  as  great  as  in  Phila- 
delphia. One-third  the  number  of  pupils  in  Chicago  minus  \ 
the  number  in  Philadelphia  was  2524.  How  many  pupils 
attended  public  high  schools  in  each  city? 

34.  The  number  of  pupils  in  public  high  schools  in  Boston 
(1905)  was  1574  less  than  4  times  as  many  as  in  Baltimore. 
Three  times  the  number  in  Boston  minus  6  times  the  number 
in  Baltimore  equals  5268.     How  many  pupils  in  each  city  ? 

35.  The  distance  from  San  Francisco  to  Yokohama  is  2031 
statute  miles  less  than  3  times  that  from  San  Francisco  to 
Honolulu,  and  the  distance  from  San  Francisco  to  Hongkong  is 
2219  more  than  twice  that  from  San  Francisco  to  Honolulu. 
One-fifth  the  distance  to  Hongkong  plus  \  the  distance  to 
Yokohama  equals  3152.  Find  the  distance  from  San  Francisco 
to  each  of  the  other  three  cities. 

36.  The  distance  from  New  York  to  Singapore  is  542  statute 
miles  more  than  3  times  that  from  New  York  to  London,  and 
the  distance  from  New  York  to  Manila  is  5342  miles  less  than 
5  times  that  from  New  York  to  London.  Three  times  the  dis- 
tance from  New  York  to  Manila  plus  6974  miles  equals  4  times 
the  distance  from  New  York  to  Singapore.  Find  the  distance 
from  New  York  to  each  of  the  other  three  cities. 


CHAPTER  II 
POSITIVE   AND   NEGATIVE   NUMBERS 

42.  Thus  far  the  numbers  used  have  been  precisely  the  same 
as  in  arithmetic,  though  their  representation  by  means  of 
letters  and  some  of  the  methods  used  in  operating  upon  them 
are  peculiar  to  algebra. 

43.  A  new  kind  of  number  will  now  be  studied  in  connection 
with  problems  of  the  following  type. 

Illustrative  Problems.  1.  If  a  man  gains' $1500,  and  then 
loses  $  800,  what  is  the  net  result  ?     Answer,  $  700  gain. 

2.  The  assets  of  a  commercial  house  are  $  250,000,  and  the 
liabilities  are  $  275,000.  What  is  the  net  financial  status  of 
the  house  ?    Answer,  $  25,000  net  liabilities. 

3.  The  thermometer  rises  18  degrees  and  then  falls  28 
degrees.  What  direct  change  in  temperature  would  produce 
the  same  result  ?    Answer,  10  degrees  fall. 

4.  A  man  travels  700  miles  east  and  then  400  miles  west. 
What  direct  journey  would  bring  him  to  the  same  final  desti- 
nation ?     Answer,  300  miles  east. 

In  the  statement  of  each  of  these  problems  the  numbers  are 
applied  to  things  which  are  opposite  in  quality ;  namely,  gain 
and  loss,  assets  and  liabilities,  rise  and  fall,  distances  measured 
in  two  opposite  directions,  as  east  and  west. 

The  answer  in  each  case  not  only  gives  the  proper  arithmeti- 
cal number,  but  also  connects  with  this  number  one  of  the  two 
opposite  qualities  involved  in  the  problem. 

E.g.  The  answer  in  (1)  is  $700  gain  and  not  simply  f  700;  in  (2) 
it  is  not  $25,000,  but  $25,000  liabilities;  in  (3)  it  is  not  10°,  but  10° 
fall ;  and  in  (4)  it  is  not  300  miles,  but  300  miles  east. 

47 


48  POSITIVE  AND  NEGATIVE  NUMBERS 

These  problems  are  illustrations  of  the  sense  in  which  the 
word  "  opposite  "  is  here  used. 

E.g.  Gain  and  loss  are  opposite  in  that  they  annul  each  other 
when  taken  together ;  rise  and  fall  of  temperature  are  opposite  in  that 
one  counteracts  the  other. 

44.  In  all  problems  involving  one  or  both  of  two  qualities 
which  are  opposite  in  this  sense  there  is  constant  need  to  dis- 
tinguish which  one  is  meant. 

E.g.  On  a  day  in  winter  it  is  not  sufficient  to  say  the  temperature 
is  5° ;  we  must  specify  whether  it  is  above  or  below  zero. 

In  the  case  of  the  thermometer  we  call  the  temperature  positive  if 
it  is  above  zero  and  negative  if  it  is  below  zero. 

These  words  positive  and  negative  are  used  to  describe  all  pairs  of 
qualities  which  are  opposite  in  the  sense  here  understood. 

The  signs  +  and  ~  stand  respectively  for  the  words  "posi- 
tive "  and  "  negative." 

E.g.  $  +  700  is  read  positive  $  700,  and  means  either  $700  gain  or  $700 
assets,  according  to  the  problem  in  which  it  occurs.  Likewise  $~700 
is  read  negative  $700,  and  means  either  $700  loss  or  $700  liability. 

Referring  to  the  thermometer,  +18°  is  read  positive  18°,  and  means 
either  a  temperature  of  18°  above  zero  or  a  rise  of  the  mercury  18°  from 
any  point.  The  latter  is  the  meaning  in  Problem  3.  Similarly  ~28° 
means  either  a  temperature  28°  below  zero  or  a,  fall  of  the  mercury  28° 
from  any  point. 

It  is  commonly  agreed  to  call  gain  positive  and  loss  negative, 
assets  positive  and  liabilities  negative,  above  zero  positive  and 
below  zero  negative,  motion  upward  positive  and  downward 
negative,  motion  to  the  right  positive  and  to  the  left  negative. 

45.  Numbers  marked  with  the  sign  +  are  called  positive  and 
those  marked  with  the  sign  ~  are  called  negative. 

The  positive  sign  maybe  omitted;  that  is,  a  number  with 
neither  sign  written  is  understood  to  be  positive. 


ADDITION  OF  SIGNED  NUMBERS  49 

ADDITION  OF  POSITIVE   AND   NEGATIVE  NUMBERS 

46.  While  in  each  of  the  problems  on  page  47  the  result 
was  obtained  by  subtracting  one  number  from  the  other, 
yet  they  are  not  properly  subtraction  problems,  but  addition 
problems. 

E.g.  In  Problem  1,  we  are  not  asking  for  the  difference  between 
$  1500  gain  and  $800  loss,  but  for  the  net  result  when  the  gain  and  the 
loss  are  taken  together ;  that  is,  the  sum  of  the  profit  and  loss.  Hence 
we  say  $1500  gain  +  $800  loss  =  $700  gain,  or,  using  positive  and 
negative  signs,  +  1500  +  ~  800  = +700.  . 

Similarly  in  Problem  2, 

$250,000  assets  +  $275,000  liabilities  =  $25,000  net  liabilities, 
Or  +  250,000  +  -  275,000  =  -  25,000. 

In  Problem  3,         18°  rise  +  28°  fall  =  10°  fall, 

Or  +18 +-28  =  -10. 

In  Problem  4, 

700  miles  east  +  400  miles  west  =  300  miles  east, 
Or  +700 +  -400  = +300. 

Note. — In  the  last  case  east  is  called  +  and  west  -,  since,  as  a  map 
is  usually  held,  east  is  to  the  right  and  west  to  the  left.  Likewise 
north  or  up  is  called  +  and  down  or  south  is  called  -. 

PROBLEMS 

1.  A  balloon  which  exerts  an  upward  pull  of  460  pounds  is 
attached  to  a  car  weighing  175  pounds.  What  is  the  net 
upward  or  downward  pull?  Express  this  as  a  problem  in 
addition,  using  positive  and  negative  numbers. 

Solution.  460  lb.  upward  pull  +  175  lb.  downward  pull  =  285  lb.  net 
upward  pull.  Using  positive  numbers  to  represent  upward  pull  and 
negative  numbers  to  represent  downward  pull,  this  equation  becomes 

+  460 +  -175  = +285. 


50  POSITIVE  AND  NEGATIVE  NUMBERS 

2.  How  would  the  morning  paper  print  the  following 
thermometer  readings  from  the  weather  report  ?  Jacksonville 
38°  above  zero,  Seattle  5°  below  zero,  Nashville  28°  above  zero, 
Chicago  13°  below  zero. 

3.  The  temperature  rises  15°  and  then  falls  24°.  What 
direct  change  in  temperature  would  produce  the  same  final 
result  ?     Express  this  as  an  example  in  addition. 

4.  A  vessel  on  the  equator  sailed  north  3°  and  was  then 
forced  south  5°  by  a  hurricane.  It  then  resumed  its  course 
northward  8°.  Express  this  as  a  sum  and  show  the  final  posi- 
tion with  reference  to  the  equator.     (See  note,  §  46.) 

In  each  of  the  following  translate  the  solution  into  the  language  of 
algehra  by  means  of  positive  and  negative  numbers,  as  in  Problem  1. 

5.  A  450-pound  weight  is  attached  to  a  balloon  which 
exerts  an  upward  pull  of  600  pounds.  What  is  the  net  up- 
ward or  downward  pull  ? 

6.  A  man's  property  amounts  to  $45,000  and  his  debts  to 
$  52,000.     What  is  his  net  debt  or  property  ? 

7.  The  assets  of  a  bankrupt  firm  amount  to  $245,000  and 
the  liabilities  to  $  325,000.  What  are  the  net  assets  or  liabili- 
ties ? 

8.  A  weight  exerting  a  downward  pull  of  280  pounds 
when  submerged  in  water  is  attached  to  a  float  which  will  just 
support  a  weight  of  240  pounds.  What  is  the  net  pressure 
upward  or  downward  when  both  are  submerged  ? 

9.  A  man  on  the  deck  of  a  steamer  is  walking  at  the  rate 
of  4  miles  an  hour  toward  the  stern.  If  the  boat  is  sailing 
eastward  in  a  river  at  the  rate  of  fifteen  miles  per  hour,  what 
is  the  actual  motion  of  the  man  with  respect  to  the  bank  of 
the  river  ? 

10.  A  man  can  row  a  boat  at  the  rate  of  6  miles  per  hour. 
How  fast  can  he  proceed  against  a  stream  flowing  at  the  rate 
of  2^  miles  per  hour ;  7  miles  per  hour  ? 


ADDITION   OF  SIGNED  NUMBERS  51 

11.  A  steamer  which  can  make  12  miles  per  hour  in  still 
water  is  running  against  a  current  flowing  15  miles  per 
hour.  How  fast  and  in  what  direction  does  the  steamer 
move  ? 

12.  A  dove  capable  of  flying  40  miles  per  hour  in  calm 
weather  is  flying  against  a  hurricane  blowing  at  the  rate  of  60 
miles  per  hour.  How  fast  and  in  what  direction  is  the  dove 
moving  ? 

13.  If  of  two  partners,  one  loses  $  1400  and  the  other  gains 
$  3700,  what  is  the  net  result  to  the  firm  ? 

14.  A  man's  income  is  $  2400  and  his  expenses  $  1500  per 
year.     What  is  his  saving  ? 

15.  A  man  loses  $  800  and  then  loses  $  600  more.  What  is 
the  combined  loss  ?  Indicate  the  result  as  the  sum  of  two 
negative  numbers. 

16.  A  man  gains  $  500  and  then  gains  $  700  more.  What 
is  the  combined  gain  ?  Express  the  result  as  the  sum  of  two 
positive  numbers. 

17.  A  tug  which  can  steam  9  miles  per  hour  in  still  water  is 
going  down  a  stream  whose  current  is  6  miles  per  hour.  How 
fast  is  the  tug  moving  ? 

18.  The  thermometer  falls  8°  and  then  17°  more.  Express 
the  combined  result  of  the  two  changes  as  a  negative  number. 

47.  Definitions.  Positive  and  negative  numbers  are  some- 
times called  signed  numbers,  because  each  such  number  consists 
of  an  arithmetic  part,  together  with  a  sign  of  quality. 

The  arithmetic  part  of  a  signed  number  is  called  its  absolute 
value. 

Thus,  the  absolute  value  of  +  3  and  also  of  ~  3  is  3.  Two 
signed  numbers  are  of  like  quality  when  they  have  the  same 
signs,  and  of  opposite  quality  when  one  is  positive  and  the 
other  negative. 


52  POSITIVE  AND  NEGATIVE  NUMBERS 

In  the  one  case  we  say  they  have  like  signs,  in  the  other  op- 
posite or  unlike  signs.  The  preceding  exercises  illustrate  the 
following  principle : 

48.  Principle  IX.  To  add  two  signed  numbers  of  like 
quality,  find  the  sum  of  their  absolute  values,  and 
prefix  to  this  the  common  sign  of  quality. 

To  add  two  signed  numbers  of  opposite  quality,  find 
the  difference  of  their  absolute  values,  and  prefix  to 
this  the  sign  of  that  one  whose  absolute  value  is  the 
greater. 

In  case  their  absolute  values  are  equal,  their  sum  is 
zero. 

The  sum  of  two  signed  numbers  thus  obtained  is  called  their 
algebraic  sum. 

49.  This  principle  may  be  further  illustrated  by  means  of 
the  following  graphic  representation  of  signed  numbers. 

On  an  unlimited  straight  line  call  some  starting  point  zero, 
and  lay  off  from  this  point  equal  divisions  of  the  line  indefi- 
nitely both  to  the  right  and  to  the  left,  as  shown  in  the  figure. 


■i     o   +i 

-HH- 


In  order  to  describe  the  position  of  any  one  of  these  division 
points,  we  require  not  only  an  integer  of  arithmetic,  to  specify 
how  far  to  count  from  the  starting  point  (the  point  marked 
zero)  in  order  to  reach  the  given  point,  but  also  a  quality  sign, 
to  indicate  the  direction  of  the  counting. 

E.g.  +  7  marks  the  division  point  7  units  to  the  right  of  zero,  and 
-  5  marks  the  point  5  units  to  the  left  of  zero.  Such  a  diagram  is 
called  the  scale  of  signed  numbers. 

The  part  of  the  scale  to  the  right,  taken  alone,  would  need  no 
signs,  and  would  picture  the  integral  numbers  of  arithmetic,  while  the 


ADDITION  BY  COUNTING  53 

two  parts  together  require  the  distinguishing  signs  of  quality,  and 
picture  the  integral  numbers  of  algebra. 

Fractions  would  of  course  be  pictured  at  points  between  the  inte- 
gral division  points,  on  the  right  or  the  left  of  the  scale,  according 
as  the  fractions  are  positive  or  negative. 

ADDITION  BY  COUNTING 

50.  In  arithmetic  two  numbers  are  added  by  starting  with 
either  and  counting  forward  as  many  units  as  there  are  in  the 
other. 

E.g.  To  add  3  and  5  we  start  with  5  on  the  number  scale  and 
count  6,  7,  8 ;  or  start  with  3  and  count  4,  5,  6,  7,  8. 

51.  In  algebra  two  signed  numbers  are  added  in  the  same 
manner  except  that  the  direction,  forward  or  backward,  in  which 
we  count,  is  determined  by  the  sign,  +  or  ~,  of  the  number 
which  we  are  adding. 

Thus,  to  add  +  7  and  +5,  begin  at  7  to  the  right  of  the  zero  point  in 
the  scale  of  signed  numbers  and  count  5  more  toward  the  right,  or 
begin  at  5  to  the  right  and  count  7  more  to  the  right,  in  either  case 
arriving  at  + 12. 

To  add  _7  and  ~5,  we  may  begin  at  7  to  the  left  and  count  5  more 
toward  the  left,  or  begin  at  5  to  the  left  and  count  7  more  in  that  di- 
rection, in  either  case  arriving  at  _  12. 

To  add  +  7  and  _  5,  begin  at  7  to  the  right  and  count  5  toward  the 
left,  or  begin  at  5  to  the  left  and  count  7  toward  the  right,  in  either 
case  arriving  at  +2. 

To  add  _  7  and  +  5,  begin  at  7  to  the  left  and  count  5  toward  the 
right,  or  begin  at  5  to  the  right  and  count  7  toward  the  left,  in  either 
case  reaching  -  2. 

I.e.  +7 +  +5  =  +12,  -7 +  -5  =  ~12,  +7 +  "5  =  +2,  ~7  +  +5  =  -2. 

52.  In  adding  several  signed  numbers,  we  reach  the  same 
result  whether  we  add  them  in  the  order  in  which  they  happen 
to  be  given,  or  in  any  other  order.     Thus  we  may  add  first  all 


54  POSITIVE  AND  NEGATIVE  NUMBERS 

the  positive  numbers  and  then  all  the  negative  numbers,  and 
finally  combine  these  two  results. 

E.g.  +  5  +  ~8  +  +7  +  -6,  taken  in  order  from  left  to  right,  gives  -3 
as  the  sum  of  the  first  two,  +4  as  the  sum  of  the  first  three,  and  ~2  as 
the  final  result.  But  +5  +  +7  s  + 12,  -8  +  "6  =  "14,  and  +12  +  -14  = 
-2,  the  same  result  as  before. 

53.  Addition  by  counting  makes  it  clear  that  two  numbers 
of  opposite  signs  tend  to  cancel  each  other  when  added. 

E.g.  In  adding  ~o  to  +8  we  start  with  +8,  which  we  reach  by  count- 
ing from  zero  eight  units  to  the  right,  and  then  add  ~5  by  counting 
five  units  to  the  left,  thus  retracing  five  of  the  units  just  counted. 
That  is,  -5  annuls  five  of  the  +8  units,  leaving  the  sum  +3. 

In  case  two  numbers  have  opposite  signs  and  equal  absolute 
values  one  completely  cancels  the  other. 
E.g.  +8 +-8  =  0. 

EXERCISES  AND  PROBLEMS 

Perform  the  following  indicated  additions : 

1.  -31 +  +42.  3.   +68 +  ~46.  5.   +104  +  ~245. 

2.  "17 +  -13.      4.  +34 +  "46.       6.  -llf  +  +8£. 
7.  +16  +  ~3  + +7  + +4  + -19.   8.  +42  +  +74  +  -92  +  -7  +  -3. 
9.  -13n  +  +7n  =  +7n  +  -13n  =  -6n. 

Solution.  By  Principle  IX,  to  add  ~13  n  to  +7  n  or  to  add  +7  n  to 
-13  n  is  to  "find  the  difference  of  their  absolute  values  and  prefix  to 
this  the  sign  of  that  one  whose  absolute  value  is  the  greater."  That  is, 
-13n  +  +7n  =  -(13  n— 7n)  =  _6n,  since  by  Principle  II,  13/i  — 7n  =  6n. 

10.   +14* +  -8*  4- "3* +  ~45*.      11.  +68x++3±x+-16x+-3x. 
12.   296r*  +  -367r*  +  -119rt.      13.    ~3  (a  +  6)  + +4  (a  +  6). 
14.    +7(x  —  y)  +  -5(x  —  y).  15.    "81  (r  +  s)  +  ~91  (r  +  s). 

16.  A  man  travels  10  miles  east,  then  23  miles  west,  and 
finally  5  miles  east.  Express  his  final  distance  and  direction 
from  the  starting  point  as  the  sum  of  three  signed  numbers. 


AVERAGES  OF  SIGNED  NUMBERS  55 

17.  The  temperature  falls  35°,  rises  24°,  falls  3°,  rises  17°, 
and  finally  falls  5°.  What  is  the  net  change  in  temperature 
between  the  last  reading  and  the  original  reading  ? 

18.  A  real  estate  firm  gains  $5000  on  one  transaction,  loses 
$2500  on  a  second,  loses  $1400  on  a  third,  and  gains  $200  on  a 
fourth.     What  is  the  aggregate  result  of  the  four  transactions  ? 

AVERAGES  OF  SIGNED  NUMBERS 

54.    Half  the  sum  of  two  numbers  is  called  their  average. 

Thus  6  is  the  average  of  4  and  8.     Similarly,  the  average  of 

three  numbers  is  one-third  of  their  sum,  and  in  general  the 

average  of  n  numbers  is  the  sum  of  the  numbers  divided  by  n. 

Find  the  average  of  each  of  the  following  sets : 

1.  10,  12,  14,  16,  18.  4.     7,    9,  11,  13,  15. 

2.  5,    9,20,30,    3.  5.     7,10,21,29,30. 

3.  11,  10,    4,    6,    5.  6.  13,    8,    9,  10,    4. 

The  average  gain  or  loss  per  year  for  a  given  number  of 
years  is  the  algebraic  sum  of  the  yearly  gains  and  losses  di- 
vided by  the  number  of  years. 

Illustrative  Problem.  A  man  lost  $400  the  first  year,  gained 
$300  the  second,  and  gained  $1000  the  third.  What  was  the 
average  loss  or  gain  ? 

Solution.     -400 +.  +  300  + +1000  = +900  =  +30Q 

3  3 

That  is,  the  average  gain  is  $  300. 

Illustrative  Problem.  During  four  years  a  certain  business 
shows  an  average  annual  gain  of  $  5000.  What  was  the  loss 
or  gain  the  first  year,  if  for  the  remaining  years  there  were 
gains  of  $8000,  $9000,  and  $7500  respectively? 

Solution.    Let  x  represent  the  number  of  dollars  gained  or  lost  the 
first  year.     Then  the  average  for  the  four  years  is 
x  +  +8000  +  +9000  +  +7500 


56  POSITIVE  AND  NEGATIVE  NUMBERS 

But  the  average  for  the  four  years  is  given  as  $  5000. 

Hence  «  +  +  8000  +  +9000  +  +  75Q0  =  +5000. 

4 

By  If,  x  +    +  8000  +    +9000  +  +7500  =  +20000. 

By  IX,  x  +  +  24500  =  +  20000. 

By  A,  x  +  +  24500  +  -  24500  =  +  20000  +  -24500. 

By  IX,  a;  =-4500. 

Hence  there  was  a  loss  of  S$  4500  the  first  year. 


PROBLEMS 

1.  Find  the  average  of  $1800  loss,  $3100  loss,  $6800  gain, 
$10,800  loss,  and  $31,700  gain. 

2.  Find  the  average  of  $180  gain,  $360  loss,  $480  loss, 
$100  gain,  $700  gain,  $400  gain,  $1300  loss,  $300  gain,  $4840 
gain,  and  $  12,000  gain. 

3.  A  merchant  gained  an  average  of  $  2800  per  year  for  5 
years.  The  first  year  he  gained  $3000,  the  second  $1500,  the 
third  $4000,  and  the  fourth  $2400.  Did  he  gain  or  lose  and 
how  much  during  the  fifth  year  ? 

4.  A  certain  business  shows  an  average  gain  of  $4000 
per  year  for  6  years.  During  the  first  5  years  the  results 
were,  $8000  loss,  $10,000  gain,  $7000  gain,  $3000  gain, 
and  $12,000  gain.  Find  the  loss  or  gain  during  the  sixth 
year. 

5.  Find  the  average  of  the  following  temperatures :  7  a.m., 
"4°;  8  a.m.,  -2°;  9  a.m.,  -1°;  10  a.m.,  +1°;  11  a.m.,  +5°; 
12  m.,  +7°. 

6.  During  the  12  hours  ending  at  6  a.m.,  January  19,  1892, 
the  U.  S.  Weather  Bureau  at  Helena,  Montana,  recorded  the 
following  temperatures:  "9°,  "8°,  "8°,  "9°,  "9°,  ~9°,  "8°,  +36°, 
+  37°,  +40°,  +20°,  +16°.  Find  the  average  temperature  for  the 
12  hours. 


AVERAGES  OF  SIGNED  NUMBERS  57 

Find  the  average  yearly  temperatures  at  the  following 
places,  the  monthly  averages  having  been  recorded  as  given 
below : 

7.  For  New  York  City:  +29°,  +33°,  +39°,  +46°,  +53°,  +63°, 
+  67°,  +67°,  +61°,  +52°,  +47°,  +41°. 

8.  For  Singapore,  Straits  Settlement :  +  81°,  +  84°,  *  85°,  +  85°, 
+85°,  +86°,  +86°,  +87°,  +85°,  +84°,  +83°,  +81°. 

9.  For  St.  Vincent,  Minnesota:   "5°,  0°,  +15°,  +35°,  +55°, 
+  60°,  +66°,  +63°,  +55°,  +40°,  +22°,  +5°. 

10.  For  Nerchinsk,  Siberia:  "23°,  -13°,  "10°,  +35°,  +55°, 
+  70°,  +70°,  +64°,  +50,  +30°,  +5°,  ~15°. 

If  the  latitude  of  a  place  is  midway  between  the  latitudes  of 
two  given  places,  then  its  latitude  equals  one-half  the  algebraic 
sum  of  the  two  given  latitudes. 

Thus,  if  the  latitude  of  one  place  is  +16°  and  that  of  the 
other  "  56°,  then  the  latitude  of  the  place  midway  between  them 

is"56°  +  +16O  =  ^==-20°. 
2  2 

The  data  used   in    some   of  the   following   problems  vary 

slightly  from  the  published  records,  but  in  no  case  more 

than  5'. 

11.  The  latitude  of  New  Orleans,  Louisiana,  is  +30°,  and  of 
Toronto,  Canada,  +43°  40'.  Find  the  latitude  of  Norfolk,  Vir- 
ginia, which  is  midway  between  these  latitudes.  (Positive 
numbers  represent  north  latitude  and  negative  numbers  repre- 
sent south  latitude.) 

12.  The  latitude  of  Alexandria,  Egypt,  is  +31°  10',  and  of 
Christiania,  Norway,  +59°  50'.  That  of  Venice,  Italy,  is  mid- 
way between  these  latitudes.     Find  the  latitude  of  Venice. 

13.  The  longitude  of  Edinburgh,  Scotland,  is  "3°  10',  and  of 
Warsaw,  Poland,  +21°.  Find  the  longitude  of  Bremen,  Ger- 
many, which  is  midway  between  these.  (Positive  numbers 
represent  east  longitude  and  negative  numbers  west  longitude.) 


58  POSITIVE  AND  NEGATIVE  NUMBERS 

14.  The  longitude  of  Vienna,  Austria,  is  +16°  20',  and  of 
Splugen  Pass,  Switzerland,  +9°  20'.  The  longitude  of  Splilgen 
Pass  is  midway  between  the  longitudes  of  Vienna  and  Paris, 
France.     Pind  the  longitude  of  Paris. 

15.  The  longitude  of  Brussels,  Belgium,  is  +4°  20',  and  of 
Jena,  Germany,  +11°  40'.  The  longitude  of  Brussels  is  mid- 
way between  those  of  Jena,  and  Liverpool,  England.  Find 
the  longitude  of  Liverpool. 

16.  The  longitude  of  Berlin,  Germany,  is  +13°  20',  and  of 
St.  Petersburg,  Russia,  +30°  20'.  The  longitude  of  Berlin  is 
midway  between  those  of  St.  Petersburg  and  Madrid,  Spain. 
Find  the  longitude  of  Madrid. 

17.  The  latitude  of  Eome,  Italy,  is  +41°  55',  and  of  Cux- 
haven,  Germany,  +53°  55'.  The  latitude  of  Rome  is  midway 
between  those  of  Cuxhaven  and  Cairo,  Egypt.  Find  the 
latitude  of  Cairo. 

18.  The  longitude  of  Calais,  France,  is  +1°  55'  and  of  Luck- 
now,  India,  +80°  55'.  The  longitude  of  Calais  is  midway  be- 
tween those  of  Lucknow  and  Washington,  D.C.  What  is 
the  longitude  of  Washington? 

SUBTRACTION  OP  SIGNED  NUMBERS 
55.   In  arithmetic  the  accuracy  of  subtraction  is  tested  by 

showing  that  the  remainder  added  to  the  subtrahend  equals 

the  minuend. 

E.g.  We  say  8  -  5  =  3  because  5  +  3  =  8. 

Indeed,  we  may,  and  often  do  perform  subtraction  by  start- 
ing with  the  subtrahend  and  counting  until  we  reach  the 
minuend. 

E.g.  A  clerk  in  changing  a  dollar  bill  after  a  purchase  of  63  cents 
might  count  out  two  pennies,  a  dime,  and  a  quarter,  saying  "65,  75, 
one  dollar."  That  is,  he  counts  from  63  cents  (the  subtrahend)  up 
to  100  cents  (the  minuend). 


SUBTRACTION  OF  SIGNED  NUMBERS  59 

56.  In  like  manner,  signed  numbers  also  may  be  subtracted, 
but  we  must  find  not  only  the  number  of  units,  but  also  the 
direction  from  the  subtrahend  to  the  minuend.  This  is  most 
easily  done  by  reference  to  the  number  scale,  page  52. 

Ex.  1.  +8  —  +5  =  +3,  since  in  passing  from  +5  to  +8  on  the 
number  scale  we  count  3  units  in  the  positive  direction. 

Ex.  2.  -8  —  -5  =  ~3,  since  from  ~5  to  "8  we  count  3  units  in 
the  negative  direction. 

Ex.  3.  "8  —  +5  =  "13,  since  from  +5  to  "8  we  count  13  units 
in  the  negative  direction. 

Ex.  4.  +8  —  ~5  =  +13,  since  from  ~5  to  +8  we  count  13  units 
in  the  j>ositive  direction. 

57.  In  each  of  these  examples  we  see  that  the  result  fulfills 
the  test  of  arithmetic ;  namely, 

Subtrahend  +  Difference  =  Minuend 

Hence  _8  —  ~5  =  -3  is  checked  by  finding  that  _5  +  -3  =  ~8. 

+8  -  -5  =  +13  is  checked  by  finding  that  "5  +  +13  =  +8. 

EXERCISES  AND  PROBLEMS 

Perform  the  following  indicated  subtractions  by  finding  the 
distance  and  direction  on  the  number  scale  from  subtrahend 
to  minuend,  and  apply  the  check  to  each  result. 

1.   -10-    "5  6.   "17 --20  11.   +93 -+22 

+6 --14  12.   +17 --13 

+7  -   -9  13.   -78  -  -37 

11-   +6  14.   +57 -+84 

-21-    "6  15.   "48 --31 

16.  How  many  degrees,  and  in  what  direction,  must  the 
temperature  change  in  order  to  vary  from  12°  below  zero  to 
38°  above  zero  ?  This  is  an  example  in  subtraction,  since  we 
are  required  to  find  a  number  which  added  to  "12°  gives  +38°. 
Hence  we  write  it  +38°  -  ~12°  =  what  ? 


2. 

-15-    +5 

7. 

3. 

+20  -  -15 

8. 

4. 

+11-   +3 

9. 

5. 

-11-   +5 

10. 

60  POSITIVE  AND  NEGATIVE  NUMBERS 

17.  How  many  degrees,  and  in  what  direction,  does  the  ther- 
mometer change  in  passing  from  27°  above  zero  to  3°  below 
zero  ?     That  is,  ~3°  -  +27°  =  what  ? 

18.  What  must  be  added  to  $35  loss  to  make  the  sum  $30 
gain  ?    That  is,  +30  -  -35  =  what  ? 

19.  What  must  be  added  to  $  15  gain  to  make  the  sum  $  8 
loss  ?    That  is,  "8  -  +15  =  what  ? 

58.  Subtraction  always  possible.  In  arithmetic  subtraction  is 
possible  only  when  the  subtrahend  is  less  than  or  equal  to  the 
minuend. 

E.g.  5  from  8  leaves  3,  5  from  5  leaves  0 ;  but  we  cannot  take  5 
from  2,  since  when  we  have  subtracted  2  units  from  2  there  are  no 
more  to  take  away. 

However,  by  means  of  negative  numbers  we  can  as  easily 
perform  the  subtraction,  2  minus  5,  as  5  minus  2. 

Thus,  2  -  5  =  -3,  since  -3  +  5  =  2;  and  5  -  2  =  +3,  since,  +3  +  2 
=  5. 

It  thus  appears  that,  in  terms  of  signed  numbers,  a  —  b  has 
a  meaning  no  matter  what  numbers  are  represented  by  a  and 
b ;  that  is,  a  —  b  means  the  number  which  added  to  b  gives  a. 

59.  A  short  rule  for  subtraction.  Since  +8  —  ~5  =  +13,  and 
since  +8  +•  +5  =  +13,  it  follows  that  subtracting  ~5  from  +8 
gives  the  same  result  as  adding  +5  to  +8.  Similarly  ~8  —  _5  = 
-3  and  "8+-+5  =  -3. 

Hence  subtracting  a  negative  number  is  equivalent  to  adding 
a  jiositive  number  of  the  same  absolute  value. 

Since  +8 -+5  = +3,  and  since  +8  +  -5  =  +3,  it  follows  that 
subtracting  +5  from  +8  gives  the  same  result  as  adding  ~5  to  +8. 
Similarly  ~8  -  +5  =  ~13  and  "8  +  ~5  =  "13. 

Hence  subtracting  a  positive  number  is  equivalent  to  adding 
a  negative  number  of  the  same  absolute  value. 


SUBTRACTION  OF  SIGNED  NUMBERS  61 

These  statements  are  illustrated  by  such  facts  as :  Removing 
a  debt  is  equivalent  to  adding  property  and  removing  property  is 
equivalent  to  adding  debt. 

Perform  the  following  subtractions  as  explained  in  this 
paragraph,  by  changing  the  sign  of  the  subtrahend  and  adding: 

1.  "5  -  -2.  5.  +57  -  -32.  9.  +37  -+50. 

2.  "4  -  +1.  6.  "32  -  +34.  10.  -23  -  +57. 

3.  -5- +2.  7.  -52 -"32.  11.  -16 a-  +4 a 

4.  +3  _  -5.  8.  "16  -  "12.  12.  +13  *  -  ~20 1 

The  preceding  exercises  illustrate  the  following  principle : 

60.  Principle  X.  To  subtract  one  signed  number  from 
another  signed  number,  add  the  subtrahend  with  its  sign 
changed  to  the  minuend. 

The  change  in  the  sign  of  the  subtrahend  may  be  made  men- 
tally without  re-writing  the  problem.  The  results  are  to  be 
checked  by  showing  that  the  difference  added  to  the  subtrahend 
equals  the  minuend. 

EXERCISES 

1.  Prom  +6  +  ~2  subtract  "14. 

2.  From  "6  a  -f  +2a  subtract  "14  a. 

3.  From  17  ab+  8  ab  subtract  "35  ab. 

4.  From  5  ax  +  4  ax  subtract  7  ax  +  2  ax. 

5.  From  54  abc  +~47  abc  -f  36  abc  subtract  80  abc. 

6.  From  54  .  13  +~47  .13  +  36-13  subtract  80  .  13. 

7.  From  29  •  3  •  11  +  37  •  3  •  11  subtract  "34  .  3  •  11. 

8.  From  29  xy  +  37  xy  subtract  ~34  xy. 

9.  Solve  x  +  8  =  4. 

Solution.  Subtract  +8  from  each  member  (which  is  equivalent  to 
adding  -8). 


62  POSITIVE  AND  NEGATIVE  NUMBERS 

Then  x  =  +4  -  +8  =  ~4,  which  is  correct,  since  ~4  +  +8  =  +4.  This 
is  a  problem  in  subtraction,  since  one  of  two  numbers,  8,  and  their 
sum,  4,  are  given,  and  we  are  to  find  the  second  number,  which  is 
represented  by  x. 


Solve  the  following  equations  : 

10.    x  +  -'3  =  7. 

16. 

x+     9=     3. 

11.   a; +  -9  =  1. 

17. 

-4  +    x  =     7. 

12.   3+   x  =  0. 

18. 

-5  +    a  =     4. 

13.   x  +  -1  =  2. 

19. 

-20+     *  =  -12. 

14.   a: +13 =7. 

20. 

8+     n=-l& 

15.   x  +   4=2. 

21. 

ft  +  -40  =  -65. 

MULTIPLICATION  OF   SIGNED   NUMBERS 
61.   The  multiplication  of  signed  numbers  is  illustrated  by 
the  following  problems. 

Illustrative  Problem.  A  balloonist,  just  before  starting,  makes 
the  following  preparations :  (a)  He  adds  9000  cubic  feet  of  gas 
with  a  lifting  power  of  75  pounds  per  thousand  cubic  feet ;  (b) 
He  takes  on  8  bags  of  sand,  each  weighing  15  pounds.  How 
does  each  of  these  operations  affect  the  buoyancy  of  the 
balloon  ? 

Solution,  (a)  A  lifting  power  of  75  lbs.  is  indicated  by  +  75,  and 
adding  such  a  power  9  times  is  indicated  by  +  9.  Hence  +  9  •  +75  = 
+  675,  the  total  lifting  power  added. 

(b)  A  weight  of  15  lbs.  is  indicated  by  ~15,  and  adding  8  such 
weights  is  indicated  by  +  8.  Since  the'  total  weight  added  is  120  lbs., 
we  have  +  8  •  "15  =  - 120. 

Illustrative  Problem.  During  the  course  of  his  journey  this 
balloonist  opens  the  valve  and  allows  2000  cubic  feet  of  gas  to 
escape,  and  later  throws  overboard  4  bags  of  sand.  What 
effect  does  each  of  these  operations  produce  on  the  balloon  ? 


MULTIPLICATION  OF  SIGNED  NUMBERS  63 

Solution,  (a)  The  gas,  being  a  lifting  power,  is  positive,  but  the 
removal  of  2000  cubic  feet  of  it  is  indicated  by  ~2,  and  the  result  is  a 
depression  of  the  balloon  by  150  lbs. ;  that  is,  _2  •  +  75  =  _150. 

(b)  The  removal  of  4  weights  is  indicated  by  ~4,  but  the  weights 
themselves  have  the  negative  quality  of  downward  pull.  Hence  to 
remove  4  weights  of  15  lbs.  each  is  equivalent  to  increasing  the 
buoyancy  of  the  balloon  by  60  lbs;  that  is,  ~4  •  _15  a  +60. 

62.  These  illustrations  of  multiplying  signed  numbers  are 
natural  extensions  of  the  process  of  multiplication  in  arith- 
metic. 

E.g.  Just  as  3  •  4  =  4  +  4  +  4  =  12,  so  3  .  -4  =  -4  +  -4  +~4  =  -12, 
and  since  3  •  4  is  the  same  as  +  3  •  +  4,  we  write  +  3.-  +  4  =  +12. 

Again,  just  as  we  take  the  multiplicand  additively  when  the 
multiplier  is  a  positive  integer,  so  we  take  it  subtractively  when 
the  multiplier  is  negative  integer. 

E.g.  -3-+4  means  to  subtract +4  three  times;  that  is,  to  sub- 
tract +  12.  But  to  subtract  +12  is  the  same  as  to  add  ~12.  Hence 
-3- +4  =~12.  Again,  -3  •  _4  means  to  subtract  -  4  three  times; 
that  is,  to  subtract  -12.  But  to  subtract -12  is  the  same  as  to  add 
+12.    Hence -3- -4  =  +12. 

EXERCISES  AND  PROBLEMS 

Explain  the  following  indicated  multiplications  and  find  the 
product  in  each  case : 


1. 

-3  •  -10. 

5. 

43  •  -192. 

9. 

71  •  -x. 

2. 

-3- +10. 

6. 

-27  •  -235. 

10. 

-112  •  ~t. 

3. 

-5  •  +50. 

7. 

~5-+r. 

11. 

-U-y. 

4. 

-75  •  -89. 

8. 

+  16- -r. 

12. 

-20  •  -v. 

13.  A  man  gained  $212  each  month  for  5  months,  then  lost 
$175  per  month  for  3  months.  Express  his  net  gain  or  loss 
as  the  sum  of  two  products. 


64  POSITIVE  AND  NEGATIVE  NUMBERS 

14.  A  man  gained  $2100  during  a  certain  year.  For  the 
first  4  months  he  lost  $125  per  month.  During  the  next  5 
months  he  gained  $500  per  month.  Find  his  gain  or  loss  dur- 
ing the  remaining  3  months  of  the  year.  Express  the  net  gain 
as  the  sum  of  two  products. 

15.  A  raft  is  made  of  cork  and  iron.  What  effects  are  pro- 
duced upon  its  floating  qualities  by  the  following  changes  ? 

(a)  Adding  4  braces,  each  weighing  (under  water)  5  lbs.  (b)  Ee- 
moving  3  pieces  of  cork,  each  capable  of  sustaining  3  lbs. 
(c)   Adding  10  pieces  of  cork,  each  capable  of  sustaining  7  lbs. 

16.  What  are  the  effects  on  a  shipwrecked  man's  ability  to 
float  ?  (a)  If  he  holds  fast  to  3  bags  of  gold,  each  weighing  1 0  lbs. 

(b)  If  he  ties  on  two  life  preservers,  each  capable  of  supporting 
15  lbs.   (c)  If  he  throws  away  his  two  boots,  each  weighing  2  lbs. 

The  preceding  exercises  illustrate  the  following  principle : 

63.  Principle  XI.  If  two  signed  numbers  are  of  the  same 
quality,  their  product  is  positive;  if  they  are  of  opposite 
quality,  their  product  is  negative.  The  absolute  value  of 
the  product  is  the  product  of  the  absolute  values  of  the 
factors. 

In  applying  this  principle  observe  that  the  sign  of  the  prod- 
uct is  obtained  quite  independently  of  the  absolute  value  of 
the  two  factors. 

E.g.  f  •  -5  =  -('£)  =  -3f ;  "12  •  ~3.5  =+42. 

64.  Principle  XI  is  also  stated  in  symbols  as  follows : 
+a  •  +6  =  +ab,  ~a  •  ~b  —+ab,  +a  •  ~b  =~ab,  ~a  •  +b  =~ab. 

The  product  of  several  signed  numbers  is  found  as  illustrated 
in  the  following : 

-2  •  +5  •  -3  •  -4  •  +6  =  -10  .  -3  •  -4  •  +6  =  +30  .  ~4  •  +6  = 
_120  •  +6  =  ~720.  That  is,  any  two  factors  are  multiplied 
together,  then  this  product  by  another  factor,  etc.,  until  all  the 
factors  are  multiplied. 


DIVISION  OF  SIGNED  NUMBERS  65 

65.  Evidently  the  factors  in  such  a  product  may  be  taken  in 
any  desired  order.  Let  the  student  try  other  orders  in  the 
above  example. 

Since  the  product  of  all  positive  factors  is  positive,  the  final 
sign  depends  upon  the  number  of  negative  factors.  If  this 
number  is  even,  the  product  is  positive ;  if  it  is  odd,  the  product 
is  negative. 

E.g.  If  there  are  5  negative  factors,  the  product  is  negative;  if 
there  are  6,  it  is  positive. 

EXERCISES 

In  the  following  exercises  determine  the  sign  of  the  product 
before  finding  its  absolute  value.  State  each  principle  used  in 
the  reduction  to  the  final  form. 

1.  -4-  +3-  ~6  .~7.  9.  «(3a+"4a-+5a). 

2.  -50  • -20  •  "30  • -40.  10.  8  (16  x  +  "20  x)  -s-  4. 

3.  ~a-  ~b  -+c-+d-+e.  11.  5x  —  3("2a;  +  +3:»— -4a>). 

4.  a>-b-+c--d--x.  12.  6r  +  4(3r  — "5r+-7r). 

5.  -5  ("3 +"7).  13.  -5  ("4  •  +3  .  -2). 

6.  a(-b  —  ~c).  14.  6x—  "14  x—  (~5  x  +~1  x). 

7.  -c{x-~y).  15.  8 y -16 y  +(4 y  +  "11  y). 

8.  1  a{x+-y—~z).  16.  lit  —  20 (* +' -3  • -5 *). 

DIVISION  OF  SIGNED  NUMBERS 

66.  In  arithmetic  we  test  the  correctness  of  division,  by 
showing  that  the  quotient  multiplied  by  the  divisor  equals  the 
dividend. 

E.g.  27  -r-  9  =  3,  because  9 -3  =  27. 

Hence  division  may  be  defined  as  the  process  of  finding  one 
of  two  factors  when  their  product  and  the  other  factor  are  given. 


66 


POSITIVE  AND  NEGATIVE  NUMBERS 


This  definition  also  applies  to  the  division  of  signed  numbers. 
In  dividing  signed  numbers,  however,  we  must  determine  the 
sign  of  the  quotient  as  well  as  its  absolute  value. 

E.g.  -42- +6  =  -7,  because  "7-  +  6  =  -42; 

also  -42  -=-  -6  =  +7,  because  +7-  -6  =  -42. 


So  in  every  case  the  test  is : 

Quotient  x  Divisor 


Dividend. 


In  like  manner  perform  the  following : 

1.  -26  +  5.  4.   -9rs-=-+3. 

2.  -ab  +  a.  5.    +  75y+~15. 

3.  +  5xy+~x.  6.    -121  x  +  +11. 

The  preceding  exercises  illustrate  the  following  principle : 

67.  Principle  XII.  The  quotient  of  two  signed  numbers 
is  positive  if  the  dividend  and  divisor  have  the  same  sign, 
negative  if  they  have  opposite  signs.  The  absolute  value  of 
the  quotient  is  the  quotient  of  the  absolute  values  of  divi- 
dend and  divisor. 

EXERCISES 

Perform  the  following  indicated  divisions.  Check  by  multi- 
plying quotient  by  divisor. 

-99  x  -=--25. 
•  87  y  ■+■  -400. 
8*-f-4y)-f--2. 
16  a  +  -20  b)  -=-  -4. 
6r  +  9s--120n--3. 
7  aa;+-14  ay-~21  &z)-=-+7. 
12  xy  —  ~3  ax)  •+■  ~x. 
27  ab  -  36  ac)  -=-  "9. 
3-4?/  +  6-8a)-f--3. 
x  —  3  y  —  z)  -f-  a. 


1. 

-28  -s-  +  7. 

11.   1 

2. 

-42  -s- -6. 

12.    1 

3. 

51  -=--17. 

13.    (< 

4. 

21xy  +  3. 

14.    (' 

5. 

"16a&-s--4. 

15.    ( 

6. 

"  15  ax  -=-  cc. 

16.    ( 

7. 

-32  (a-  6)  + 

_(a- 

-&). 

17.    C 

8. 

21(x  +  y)  + 

9. 

18.    (! 

9. 

4.-9z-=--3. 

19.    (. 

10. 

3-8*-s--4 

20.    2 

INTERPRETATION  OF  SIGNED  NUMBERS  67 

68.  While  Principles  I-VIII  were  studied  in  connection 
with  unsigned,  or  arithmetic  numbers  only,  it  is  now  very 
important  to  note  that  they  all  apply  to  signed  numbers  as 
well.  The  form  changes  described  in  §  37  also  apply  to  signed 
numbers  just  the  same  as  to  arithmetic  numbers. 

In  the  statement  of  these  principles  the  word  number  will 
from  now  on  be  understood  to  refer  either  to  the  ordinary  num- 
bers of  arithmetic  or  to  the  signed  numbers,  as  occasion  may 
require.  It  should  also  be  noticed  that  the  numbers  of  arith- 
metic are  used  as  freely  in  algebra  as  in  arithmetic.  It  is  only 
when  we  wish  to  distinguish  them  from  negative  numbers  that 
they  are  called  positive  numbers. 

The  number  system  of  algebra,  as  far  as  we  have  studied  it, 
consists  of  the  numbers  of  arithmetic  together  with  the  negative 
numbers. 

INTERPRETATION  AND  USE  OF  NEGATIVE  NUMBERS 

69.  Illustrative  Problem.  Divide  34  units  into  two  parts  such 
that  one  part  is  equal  to  the  remainder  when  3  times  the  other 
part  is  subtracted  from  46. 

Solution.    Let  the  two  numbers  be  represented  by  x  and  34  —  x. 

Then                                     34-x  =  46-3x.  (1) 

S|34,                                    -x  =  12-3z.  (2) 

,4|3ar,                                8*-x=12.  (3) 

Principle  II,                           2  x  =  12.  (4) 

D\2,                                          x  =  Q.  (5) 

Substituting  x  =  6  in  (1),    34  -  6  =  46  -  18,  or  28  =  28. 

In  the  above  solution,  equation  (2)  was  obtained  from  (1)  by  sub- 
tracting 34  from  each  member.  This  would  clearly  be  impossible 
without  the  use  of  negative  numbers. 

In  this  case  the  problem  itself  does  not  involve  negative  numbers, 
but  in  the  course  of  its  solution  they  naturally  occur.  If  the  nega- 
tive number  could  not  be  used,  we  should  be  compelled  to  keep  the 


68  POSITIVE  AND  NEGATIVE  NUMBERS 

members  of  each  equation  positive  or  zero.  This  would  be  impossible, 
since  we  do  not  know  what  numbers  are  represented  by  the  letters 
involved,  and  hence  cannot  tell  by  inspection  whether  a  given  term 
is  positive  or  not. 

70.  We  have  seen  how  naturally  the  use  of  signed  numbers 
has  arisen  in  problems  where  things  of  opposite  qualities  have 
to  be  distinguished. 

In  solving  a  problem,  a  negative  result  may  have  a  natural 
interpretation  or  it  may  indicate  that  the  conditions  of  the 
problem  are  impossible. 

A  similar  statement  holds  in  reference  to  fractional  answers  in 
arithmetic.  For  example,  if  we  say  there  are  twice  as  many  girls  as 
boys  in  a  schoolroom  and  35  pupils  in  all,  the  number  of  boys  would 
be  35  -^-  3  =  llf,  which  indicates  that  the  conditions  of  the  problem 
are  impossible. 

71.  Illustrative  Problem.  The  crews  on  three  steamers  to- 
gether number  94  men.  The  second  has  40  more  than  the  first, 
and  the  third  20  more  than  the  second.  How  many  men  in 
each  crew  ? 

Solution.         Let  n  =  number  of  men  in  first  crew. 

Then  n  +  40  =  number  of  men  in  second  crew, 

and  n  +  40  +  20  =  number  of  men  in  third  crew. 

Hence  n  +  n  +  40  +  n  +  40  +  20  =  94, 

and  3  n  +  100  =  94. 

n  s=  -2. 

Here  the  negative  result  indicates  that  the  conditions  of  the 
problem  are  impossible. 

72.  Illustrative  Problem.  A  real  estate  agent  gained  $  8400  on 
four  transactions.  On  the  first  he  gained  $6400,  on  the  second 
he  lost  $  2100,  on  the  third  he  gained  $  5000.  Did  he  lose  or 
gain  on  the  fourth  transaction  ? 


INTERPRETATION   OF  SIGNED  NUMBERS  69 

Solution.  Since  we  do  not  know  whether  he  gained  or  lost  on  that 
transaction,  we  represent  the  unknown  number  by  n,  which  may  be 
positive  or  negative,  as  will  be  determined  by  the  solution  of  the 
problem. 

Thus  we  have  6400  +  -2100  +  5000  +  n  =  8400.  (1) 

Hence  by  IX,  F,  +  9300  +  n  =  8400.  (2) 

By  S,  n  =  8400  -  9300.  (3) 

ByX,  n  =  -900.  (4) 

In  this  case  the  negative  result  indicates  that  there  was  a  loss  on 
the  fourth  transaction. 

PROBLEMS 

In  the  following  problems  give  the  solutions  in  full  and  state  all 
principles  used,  together  with  the  interpretation  of  the  results : 

1.  A  man  gains  $2100  during  one  year.  During  the  first 
three  months  he  loses  $125  per  month,  then  gains  $500  per 
month  during  the  next  five  months.  What  is  the  gain  or  loss 
per  month  during  the  remaining  four  months  ? 

2.  A  man  gained  $  1250  during  four  months.  During  the 
second  month  he  gained  $600  more  than  the  first  month,  the 
third  month  he  gained  $300  less  than  the  second,  and  the 
fourth  he  gained  $200  more  than  the  third.  Find  the  gain 
or  loss  for  the  first  month. 

3.  A  box  containing  a  Christmas  toy  weighed  25  oz.  When 
the  toy  and  the  packing  were  removed,  the  box  weighed  20  oz. 
The  packing  weighed  7  oz.    What  kind  of  a  toy  was  it  ? 

4.  A  man  rowing  against  a  swift  current  rows  8  miles  in  5 
hours.  The  second  hour  he  rows  one  mile  less  than  the  first, 
the  third  two,  miles  more  than  the  second,  and  the  fourth  and 
fifth  one  mile  more  each  than  he  rowed  the  third  hour.  How 
many  miles  did  he  row  each  hour  ? 

5.  There  are  three  trees  the  sum  of  whose  heights  is  108 
feet.  The  second  is  40  feet  taller  than  the  first,  and  the 
third  is  30  feet  taller  than  the  second.     How  tall  is  each  tree  ? 


70  POSITIVE  AND  NEGATIVE  NUMBERS 

Find  the  average  yearly  temperature  at  each  of  the  follow- 
ing places,  the  average  monthly  temperatures  being  as  here 
given  : 

6.  Port  Conger,  off  the  northwest  coast  of  Greenland ;     37°, 
-43°, "32°, -15°, +14°, +18°,  +35°,  +34°,  +25°,  +4°,  -17°,  - 30°. 

7.  Franz  Joseph's  Land ;  "20°,  "20°,  "10°,  0°,  15°,  30°,  35°, 
30°,  20°,  10°,  0°,  "10°. 

8.  Western  Baffin  Land ;  "30°,  "30°,  ~20°,  0°,  20°,  35°,  40°, 
35°,  25°,  10°,  -10°,  -20°. 

Find  the  average  yearly  loss  or  gain  in  each  of  the  following : 

9.  $1600  gain,  $8000  loss,  $24,000  gain,  $40,000  loss. 

10.  $32,000  gain,  $45,000  loss,  $24,000  gain,  $42,000  loss. 

11.  The  average  yearly  temperature  of  north  central  Siberia 
is  ~5°.  The  average  monthly  temperatures  beginning  in  Febru- 
ary are:  "60°,  "30°,  0°,  15°,  40°,  40°,  35°,  30°,  0°,  "30°,  -50°. 
Find  the  temperature  for  January. 

12.  The  business  transactions  of  a  certain  firm  averaged 
$  1500  loss  for  4  years.  For  the  first  year  there  was  a  gain  of 
$  800,  the  second  year  a  loss  of  $  1800,  the  third  year  a  loss  of 
$300.     What  was  the  loss  or  gain  for  the  fourth  year  ? 

13.  A  commercial  house  averaged  $15,000  gain  for  5 
years.  What  was  the  loss  or  gain  the  first  year  if  the  remain- 
ing years  show :  $8000  gain,  $24,000  gain,  $2000  loss,  $20,000 
gain,  and  $  50,000  gain,  respectively  ? 

14.  The  longitude  of  Boston,  Massachusetts,  is  "71°  10',  and 
that  of  Chicago,  Illinois,  is  ~87°  35'.  Find  the  longitude  of 
Lake  Chautauqua,  which  is  midway  between  these. 

15.  The  longitude  of  New  Haven,  Connecticut,  is  ~72°  58', 
and  that  of  Bombay,  India,  is  +72°  48'.  The  longitude  of  St. 
Paul's  Cathedral,  London,  is  midway  between  these.  Find  the 
longitude  of  the  cathedral. 


INTERPRETATION  OF  SIGNED  NUMBERS  71 

16.  The  longitude  of  Cincinnati,  Ohio,  is  ~84°  30',  and  that 
of  Indianapolis,  Indiana,  is  "86°  5'.  The  longitude  of  Cincin- 
nati is  midway  between  those  of  Indianapolis,  and  Columbus, 
Ohio.     Find  the  longitude  of  Columbus. 

17.  The  longitude  of  Bristol,  England,  is  ~2°  30',  and  that  of 
Minneapolis,  Minnesota,  is  ~93°  20'.  The  longitude  of  Bristol 
is  midway  between  those  of  Minneapolis  and  Calcutta,  India. 
Find  the  longitude  of  Calcutta. 

18.  The  latitude  of  Columbus,  Ohio,  is  +40°  and  that  of 
Winnipeg,  Canada,  is  +50°.  The  latitude  of  Columbus  is  mid- 
way between  those  of  Winnipeg  and  Houston,  Texas.  Find 
the  latitude  of  Houston. 

19.  The  longitude  of  Montreal,  Canada,  is  "73°  40'  and  that 
of  Baltimore,  Maryland,  "76°  40'.  Find  the  latitude  of  Phila- 
delphia, which  is  midway  between  these. 

20.  The  latitude  of  Lima,  Peru,  is  ~12°  and  that  of  Buenos 
Ay  res,  Argentina,  "34°  35'.  The  latitude  of  Lima  is  midway 
between  those  of  Buenos  Ayres  and  Caracas,  Venezuela.  Find 
the  latitude  of  Caracas. 

21.  The  longitude  of  Providence,  Rhode  Island,  is  _71°  75' 
and  that  of  Fargo,  North  Dakota,  -96°  50'.  The  longitude  of 
Fargo  is  midway  between  those  of  Providence  and  Seattle, 
Washington.     Find  the  longitude  of  Seattle. 


I 


CHAPTER  III 
INVOLVED  NUMBER  EXPRESSIONS 

73.  Double  Use  of  the  Signs  +  and—.  In  the  preceding 
chapter  it  has  been  found  that  the  negative  quality  may  be 
regarded  as  implying  subtraction  and  the  positive  quality  as 
implying  addition.  It  was  for  this  reason  that  +  and  ~  were 
selected  as  symbols  for  the  words  "  positive  "  and  "  negative." 

74.  It  is  now  possible  to  dispense  with  these  special  signs 
of  quality.  For,  by  Principle  X,  a  —  +b  and  a  +  ~  b  are  the  same 
in  effect,  and  likewise  a—~b  and  a-\-+b.  Hence,  omitting 
the  positive  signs  (§  45),  we  may  write 

a  -f-  ~&  =  a  —  b  and  a  —  ~b  =  a  +  b. 

One  set  of  signs  is,  therefore,  sufficient  as  symbols  both  of 
operation  and  of  quality. 

E.g.  5  —  7  means  either  5  +~7  or  5  —  +7,  and  in  either  case  equals 
~2,  which  we  now  write  —  2.  Thus  5  —  7  equals  —  2,  which  is  read, 
5  minus  7  equals  negative  2. 

Ex.  1.  "5.-4  =  +  20  is  now  written  -  5  •  -  4,  or  (—  5)(-4), 
=  +  20,  V_    ' 

and      -5-+4  =  -20is  written -5  • +4  or,  (-5)  (  +  4),= -20. 

Ex.2.  5(9  a +"2  a)  =  5(9  a— 2  a). 

By  IV,  =45  a- 10a. 

By  II,  =  35  a. 

Ex.3.  5a+-4(-3a-+76)  =  5a-4(-3a-7&). 
By  IV,  XI,  =5a  +  12a  +  28&. 

By  I,  =17  a +28  6. 

72 


POLYNOMIALS  73 

EXERCISES 

Rewrite  the  following,  using  one  set  of  signs,  and  then  per- 
form the  indicated  operations. 

1.  7- +3 +  "8.  5.  27  abc +  ~35abc- 2  abc. 

2.  -9 --3 +  -12.  6.  5 ("2  +  -3)  +  4(5  -~7). 

3.  -4a  +  -5a  +  "6a.  7.  '8 (7  -  2)  -  8(6  +  "9). 

4.  -3-5a;  +  4.7a;-8.8;K.  8.  3(4a-"5  b)  -11(~2  a+~36) 

Rewrite  the  following  expressions,  using  special  signs  of 
quality  so  that  all  signs  of  operation  shall  indicate  addition  : 

9.  5-8-14  =  5 +-8 +-14.  13.  56 ay  —  72 ay  +  7ay. 

10.  -7  +  8-18.  14.  3(2-6)  +  6(3-7). 

11.  -4a  +  5a-17a.  15.  -3(8  -  6) -4(6-9). 

12.  -7-4x+7-4x--  8  -4a:.  16.  8(4*  -5?t)-5(—  t  +  4n). 

POLYNOMIALS 

75.  We  have  found  that  the  solution  of  problems  leads 
us  to  build  involved  number  expressions  out  of  single  number 
symbols. 

E.g.  If  a;  is  a  number  representing  my  age  in  years,  then  2 (x  —  10) 
is  double  the  number  representing  my  age  10  years  ago,  and 
2  l(x  —  10)  +  (x+ 15)]  is  the  number  representing  twice  the  sum  of  my 
ages  10  years  ago  and  15  years  hence. 

Number  expressions  are  now  to  be  studied  more  in  detail. 

76.  Definition.  A  number  expression  composed  of  parts  con- 
nected by  the  signs  +  and  —  is  called  a  polynomial.  Each  of 
the  parts  thus  connected  together  with  the  sign  preceding  it  is 
called  a  term. 

E.g.  5  a  -  3  xy  —  f  rt  +  99  is  a  polynomial  whose  terms  are  5  a, 
-Zxy,  -  |  rt,  and  +  99.     The  sign  +  is  understood  before  5  a. 


74  INVOLVED   NUMBER   EXPRESSIONS 

77.  Definitions.  A  polynomial  of  two  terms  is  called  a  bino- 
mial, one  of  three  terms  is  called  a  trinomial.  A  term  taken 
by  itself  is  called  a  monomial. 

E.g.  5  a  —  3  xy  is  a  binomial;  5  a  —  3  xy  —  §  rt  is  a  trinomial ;  5  a, 
—  3  xy,  —%rt  are  monomials. 

According  to  the  above  definition  x  +  (6  +  c)  may  be  called 
a  binomial  notwithstanding  it  is  equivalent  to  the  trinomial 
x  +  b  +  c. 

In  this  case  x  is  called  a  simple  term  and  (b  +  c)  a  compound 
term.  Likewise  we  may  call  3t  +  £x  —  5(a-\- b)y  a  trinomial 
having  the  simple  terms  3  t,  -f  4  x,  and  the  compound  term 
-5(a  +  b)y. 

It  should  be  clearly  understood  that  a  negative  or  positive 
sign  before  a  compound  term  (as  well  as  before  a  simple  term) 
applies  to  the  number  represented  by  the  whole  term. 

78.  Definition.  Two  terms  which  have  a  factor  in  common 
are  called  similar  with  respect  to  that  factor. 

E.g.  5  a  and  —  3  a  are  similar  with  respect  to  a ;  —  3  xy  and  —  7  x 
are  similar  with  respect  to  x ;  5  a  and  —  5  b  are  similar  with  respect 
to  5 ;  7  abc  and  —  f  abc  are  similar  with  respect  to  abc. 

Similar  terms  may  be  combined  by  Principles  I,  II,  and  IX. 

E.g.  5a-3a  =  (5-3)a  =  2a;  -3xy-7x=  -x  (3  y+7)  ;  5a-5b  = 
5(a-b). 

ADDITION  AND  SUBTRACTION  OF  POLYNOMIALS 

79.  In  adding  or  subtracting  polynomials  the  work  may  be 
conveniently  arranged  by  placing  the  terms  in  columns,  each 
column  consisting  of  terms  which  are  similar. 

Ex.  Add  5x-6y  +  4:z  +  5at,  —  3  x  +  11  y  -16z-  96*, 
and  —7y-\-8z. 


ADDITION  OF  POLYNOMIALS  75 

Arranging  as  suggested  and  applying  Principles  I,  II,  and  IX, 

we  have 

5x—   6y  +   4  2  +  5  at 

-3x  +  lly-16z-9bt 

-  iy+  8z 

2x  —   2y-  4:Z  +  t(5a-9b) 

5  x  and  —  3  x  are  similar  with  respect  to  their  common  factor 
x.  Hence  by  Principle  I  we  add  the  other  factors  5  and  —  3, 
obtaining  (5  —  S)x  =  2x. 

Likewise  we  add  +  5  at  and  —  9  bt  with  respect  to  the  com- 
mon factor  t,  obtaining  (5  a  —  9  b)t.  In  the  second  column  the 
sum  is  (—  6  +  11  —  l)y=  —  2y,  and  in  the  third  column  the 
sum  is  (  +  4  —  16  +  8) z  =  —  4 z. 

Check  by  giving  convenient  values  to  x,  y,  z,  t,  a,  and  b. 

EXERCISES 

1.  Add76-3c  +  2o';  _26  +  8c-13d\ 

2.  Add  6x—  3  y  +  ±t  —  7  z;   x  —  5  y-3t;  4  x-±y  +  8  t. 

3.  Add  5  ac +  3  6c -4c  +  86;  26  +  3c-2  6c -3 ac;  46  + 
4 c  +  6c  —  ac;  2  6c  +  4ac  +  c;  36— 4c. 

4.  Add  3-4.7  —  5x  +  5abx;  3aby  +  3x  —  5  -4  ■  7;  7 x  + 
2-4-7;  baby  —  3x  —  5-4-  7;  45#  +  a6cc  —  4  -7. 

5.  Add  3(x  —  5)  +  a(c  +  6)  +  6(a-?/);  6(c  +  6)  —  a(x  —  y) 
+  8(a>-5);  7(c  +  6)-4(a-  </);  3(a>-y)  +  (a-5). 

6.  Add  16(a+6-c)-3(x-2/)+2(a-6);  2{x-y)- 
3(a-6)  +  (a+  6— c);  7 (a  —  6)  +  a(x  —  y)  —  6(a  +  6  —  c). 

7.  Add  a(a  —  6)  —  c(#  +  ?/)  +  d(x  —  z)—  4a6c;  c(#  —  2)— 
d(x+y)  +  (a— 6)+2a6c;  e(a  — 6)+ma6c+3(a  — z)+8(a+y). 

8.  Add  7a  —  4a+12  2;  6a  —  3  x  +  C2 ;  2  6a  +  4  a  —  3  cz. 


76  INVOLVED  NUMBER  EXPRESSIONS 

9.  Add   (a  —  &)  —  3(c  —  d)+m(a  +  d) ;   c(a—b)+a(c  —  d). 

10.  Add  34  ax  +  4  by  — 3  z;  2by  +  5z;  3bx  —  7y  +  5dz. 

11.  Add   3b  +  4:cd—2ae;    ab—3cd  +  3ae;   3cd  —  2ab. 

12.  Add  7ax-13by  +  5;  9ax  +  8by-4;  3  b -12 ax. 

13.  Add5(a  +  6)-3(c-d);  3(c-d)-8(a  +  6);  — 2(a  +  6). 

14.  Add  3  +  4(c  -d)-5(a-6-c)  ;  4 (a— 6  -  c)  +  5(c-d); 
3(a_6_c)-9(c-d)  +  12. 

15.  Add  U(c  —  9)  +  3(x  +  y)  +  21ivu;  —71wu  —  5(x  +  y) 
-13(c-9). 

16.  Add  5a&  -  3- 7 -9  + 5(a-l);  5- 9- 7 +  3a&- 2  (z-1); 
3(»-l)-4.7.9  +  2a&. 

17.  Add  31  •  50  -  43  •  74  +  2  •  18 ;  21  •  74  +  7  •  18  -  56  •  50  ; 
-  12  •  18  +  42  •  50  -  6  •  74. 

18.  Add  7(a5-y)-4(aj  +  y)  +  4-7;  9  (a;  4-  2/)  +  3(x-y)  - 
9-7;  6(^-^)4-2.7-3(^  +  2/). 

19.  Add  16  xy  - 13  •  64 ;  15  a&  -  2  a# ;  34  •  64  -  3  xy  4-  2  a& ; 
14- 64 -3a?/- 2 a&. 

80.  The  subtraction  of  polynomials  is  illustrated  by  the 
following  example : 

From  15  ab  — 17  xy  + 11  rt  subtract  —  5  ab  4-  4  xy  —  5  nt. 

Arranging  as  on  page  75  and  applying  Principles  II  and  X, 

15  ab  —17xy  +  11  rt 
—  5ab  +    4  xy  —    5  n< 
20 ab -21  xy+  t(Ur  +  5  n) 

As  suggested  in  §  60,  it  is  sufficient  to  change  the  signs  of  the  sub- 
trahend mentally,  rather  than  to  rewrite  them  before  adding  to  the 
minuend. 


SUBTRACTION   OF  POLYNOMIALS  77 

EXERCISES 

1.  From  9x+3y— 11  z  subtract  —  5x+8y—  3z. 

2.  From  12  ab  —  3  cd  + 12  xy  subtract  3  ab  +  2  cd  —  11  xy. 

3.  From  9  xc  +  4  ad  —  3  cz  +  2  ?/  subtract  3  ?/  —  3  ad  +  5  cz. 

4.  From  13  t  +  5  mx  —  5  cu   subtract  2  £  —  4  ma  +  3  cv. 

5.  From  3  v  —  2  w  +  5  ?nn  —  4  asz  subtract  —  -y  +  5  w  —  3  raw. 

6.  From  31  b  +  4  ary  -f  10  a.c  —  4  subtract  8  6  —  5  xy  —  3  ax. 

7.  From  4  —  3  a  —  5xz—  3vy-x  subtract  7  a  +  2  az  4-  4  vy. 

8.  From  8  .ry  —  3  x  +  4  y  subtract  —  2xy  +  13w  +  4x—2y. 

9.  From  2  a&  —  5  +  7  v  + 13  abc  subtract  3  ab  +  v  +  8  a&c. 

10.  From  8 ca;a  —  4yb—3yc  subtract 4 fom + 2 yb  +  4 yc  —  49. 

11.  From  31  •  ir>  —  7xy   subtract    12  •  45 +  9  a*/. 

12.  From   3  abc  -  4 -28  + 2  (x  +  y) -3  xy    subtract    6  •  28 
+  4  xy  —  3  (x  +  y)  +  8  a&c. 

13.  From  7  •  3  •  5  +  9  (»y  —  z)  -f  4  •  3  (a  +  6)  subtract  8(a#  —  z) 
-8-3(a  +  b)+8-3-5. 

14.  From  5  ax  — 3  by +  4  ax+  5  by  subtract  5  by— 3ax+  7  by. 

15.  From     19  (r  -  5  s)  +  13  (5  x  —  4)  +  7  (a?  —  y)     subtract 
17(5  x  _  4) -  5  (x-  y) - 11  (r  -  5  s). 

16.  From   16  - 15  •  30  + 14(*  -  5  yz)  — 13  (5  y  —  z)    subtract 
32-16- 30 +  8(5y-*). 

17.  From  -41.3  +  13-4-16  subtract   7-  4-  16-8-3. 

18.  From      a  (b  +  c)  +  4  (ra  +  »)  —16  c     subtract     9  (ra  +  n) 
+  31c-d(6  +  c). 

19.  From    5(7a;-4)  +  3(5t/-3a;)  +  5  -  7   subtract    8-7 
-9(7  x  -  4) + 8(5  y— 3*). 

20.  From  15  •  48  +  8  a&  +  49  a;  subtract  7  •  48  -  9  a&  - 14  x. 


78  INVOLVED  NUMBER  EXPRESSIONS 

EXERCISES  IN  ADDITION  AND  SUBTRACTION 

1.  Add  5x  —  3y  —  7  r  +  8t,  —7  x  +  18y  —  ±r  —  7t,  —20  a; 

—  24  y  + 18  r  - 15 1,  and  13  x  + 15  y  +  11  r  +  6  t. 

Check  the  sum  by  substituting  a;  =  1,  y  =  1,  r  =  1,  f  as  1. 

2.  Add  17a  — 9  6,  3c  +  14a,  6  — 3  a,  a  — 17c,  and  a  —  36 
-f  4  c.     Check  for  a  =  1,  6  =  2,  c  =  3. 

3.  Add  2x  +  3y  —  t,  —6y  +  8t,  —x  +  y  —  t,  —  4*  +  7z,and 
3  y.     Check  f or  x  =  2,  y  =  3,  £  =  1. 

4.  Add  17r  +  4s-£,  2t  +  3u,  2r-3s-f4£,  5w-6«,  7r 

—  3s  +  8u,  and  8r  — 2f  +  6w.     Check  by  putting  each  letter 
equal  to  1 ;  also  equal  to  2. 

5.  Add  3  ft  +  2  £  +  4  u  and  ft  +  3 1  +  3  w.     Check  by  putting 
ft  =  100,  t  =  10,  rt  =  1 ;  i.e.  324  + 133  =  457. 

6.  Add  4  ft  +  3  £  +  u  and  3  ft  +  2  £  -f  7  u.     Check  as  in  5. 

7.  Write  247,  323,  647,  239,  and  41,  as  number  expressions 
like  those  in  5  and  6  and  then  add  them. 

8.  Add  At  —  u,  bt  —  u,  6t  —  u,  7  t  —  u,  and  8 1  —  u.     Check 
for  t  =  10,  u  =  1 ;  also  t  =  1,  u  =  1. 

9.  Add  647,  391,  276,  and  444  as  in  example  7. 

10.  Simplify:  3 xyz  —  2 xyz  +  5 xyz  —  4 xyz  +  xyz  —  xyz. 

11.  Subtract  5 a  —  3 6  +  6 c  from  —  8a-f  76  — lie  and  check. 

12.  From  7  xy  +  8  xz  +  9  yz  take  17  a;y  — 19  xz  —  20  yz. 

13.  From  6  a;  — 3  y  take  8  y  — 3  z. 

14.  From  3p  —  4  g  +  8  r  take  7j9  — 11  r  + 11 5- 

15.  From  a  +  b  +  c  take  a;  —  y  +  z.    Suggestion :  By  Principle 
VII,  a  +  6  +  c-(a;  —  y  +  z)=za  +  b  +  c-x  +  y  —  z. 

16.  From  2a;  — 3y  take  5. r +  7 y  + 2 a  — 3  6. 

17.  From  the  sum  of  18  a6c  —  27  xyz  + 13  rst  and  — 11  a6c 
-f 16  xyz  —  52  rst  take  67  rst  —  39  a6c. 


ADDITION  AND   SUBTRACTION  79 

18.  To  the  difference  between  the  subtrahend  15  a;  — 18  y 
+  27  z  and  the  minuend  117  x  +  07y  —  81  z  add  4  a:  —  6 y  +  3z. 

19.  Add  ll(a?-y)  + 15  (a  -6)  and  -20(a5-y)-37(o-6) 
and  from  the  sum  subtract  135  (a;  —  y)—  213  (a  —  b). 

20.  Add  6ax  +7  bx  —  8cx,  —11  ax  — 18  &#  +  25  ex,  and  19  ax 
-16cx  +  24:bx. 

21.  From   13  mn  —  25  mp  +  36  mq   subtract   18  mn  +  23  mp. 

22.  Add  by  Principle  I,  6  •  3  •  9  - 11  •  5  •  7  + 16  ■  9  •  11  and 
-  8  •  3  .  9  +  24  •  5  •  7  -  23  •  9  .  11. 

23.  From  83-9  +  78 -13   subtract   57-9-93.13+85.17. 

24.  From   3 h  +  4 1  +  2 u  subtract  h  +  5t+3u.      Check. 

25.  Subtract  7(a— x)— 10 (b  —  y)  from  13(a  —  x)+5(b  —  y). 

81.  In  solving  problems  it  is  often  unnecessary  to  arrange  the 
work  of  addition  and  subtraction  in  columns  as  above.  In  most 
cases  the  operations  can  be  readily  indicated  by  means  of  paren- 
theses as  illustrated  in  the  following  example : 

From  the  sum  of  3  a  +  4  b  and  5  a  — Sb  subtract  7  a  — 6  b. 

Indicate  these  operations  thus  :  3  a  +  4  b  +  (5  a  —  8  6)  —  (7  a  —  6  b). 
Applying  Principle  VII,  we  have  3a  +  46  +  5a  —  86  —  7 a  +  Qb. 
Collecting  similar  terms,  3a  +  5a  —  7a  +  46  —  86  +  6  6. 
Finally,  applying  Principles  I  and  II,  we  obtain  a  +  2  6. 
After  a  little  practice  the  last  two  steps  can  be  taken  at  once. 

EXERCISES  AND  PROBLEMS 

Perform  the  following  indicated  operations  by  collecting 
similar  terms  at  once  without  arranging  in  columns : 

1.  15 +  (7 -9 a;)- (-7* -9) +9. 

2.  7  +  5y-(3y  +  2)  +  (8-4:y). 

3.  2a  +  3  +  (4a-5)-(lla-14). 

4.  326-(176-12)-(46-13). 


80  INVOLVED  NUMBER   EXPRESSIONS 

5.  16c-(41-7c)  +  (15-8c). 

6.  -  (5  a  -  3  c)  -  (2  c  -  8  a)  +  3  a. 

7.  _(_  12  a  _7 ?/ -15  a:) -(-9 # +  8  a +  3?/). 

8.  (19  x  +  4 y  -  32  x  - 17  x)  - 12  a  -  (49  y  + 18 a>  -  70  a). 

9.  17a-3-(7a-2)  +  (6a-5). 

10.  5a-  (8-4x  +  7y)  +  (5x  +  3)-(5y  +  3x-99). 

11.  -(3a  +  56-7c)  +  (8a-4c)-(9c-46  +  4a)-91a. 

12.  7-(4-4c  +  2d-2a)+31c-(4-2a-5d)-(-8c). 

13.  (41a&-21c+4)-(36c  +  15-78a6)  +  (13c-90ao-8). 

14.  9  by  -  (4  c  -  8  by  - 13)  -  2c  - 16  -  (34  by -12  c  +  8  by). 

15.  6 mn+(—  9m  —  7 w  +14) -8 n  +  (13 mn—11  m)+34 raw. 

16.  34 ax—  (— 17 a#  + 42) +  8 a -(14 a +  24 ax  —  7). 

17.  19  -  (+  2  -  7a  -  4  b  +  11  oft)  -  (-  2  6  +  8  ab  +  4  a). 

18.  41  %  -  (4  b  - 13  y  + 17  by)  -  (-  5  b  - 17  6y  + 13  y). 

19.  39  rs  -  20  s  - 19  r  -  (7  »•*  +  8  s  - 19  r)  -  (15  r  -  5  s  -  56). 

20.  a(3  x  —  2  y  —  z)  —  (5  ax  —  ay  +  3  z)  +  az. 

21.  5(4  h  +  3bk-7br)-  6(15  &  -  35  r)  -  20  h. 

22.  The  altitude  of  Popocatepetl  is  1716  feet  less  than  that 
of  Mt.  Logan,  and  the  altitude  of  Mt.  St.  Elias  is  316  feet 
greater  than  that  of  Popocatepetl.  Find  the  altitude  of  each 
mountain,  the  sum  of  their  altitudes  being  55,384  feet. 

23.  The  Ganges  River  is  1800  miles  shorter  than  the  Ama- 
zon, and  the  Orinoco  is  300  miles  shorter  than  the  Ganges. 
The  sum  of  their  lengths  is  6900  miles.     How  long  is  each  ? 

24.  A  cubic  foot  of  red  oak  weighs  35  pounds  less  than  2 
cubic  feet  of  cherry  wood,  and  21  pounds  more  than  a  cubic 
foot  of  chestnut ;  while  a  cubic  foot  of  chestnut  weighs  100 
pounds  less  than  3  cubic  feet  of  cherry.  Find  the  weight  of 
each  kind  of  wood  per  cubic  foot. 


ADDITION  AND   SUBTRACTION  81 

25.  Lead  weighs  259  pounds  more  per  cubic  foot  than  cast 
iron,  and  166  pounds  more  than  bronze ;  while  a  cubic  foot  of 
bronze  weighs  807  pounds  less  than  3  cubic  feet  of  iron.  Find 
the  weight  per  cubic  foot  of  each  metal. 

26.  Green  glass  weighs  60  pounds  per  cubic  foot  less  than 
dense  flint  glass,  and  8  pounds  more  than  crown  glass ;  while  a 
cubic  foot  of  crown  glass  weighs  293  pounds  less  than  2  cubic 
feet  of  flint  glass.    Find  the  weight  per  cubic  foot  of  each. 

27.  Europe  has  12  million  inhabitants  less  than  10  times  as 
many  as  South  America,  and  North  America  has  29  million 
more  than  twice  as  many  as  South  America.  If  3  times  the 
population  of  North  America  be  subtracted  from  that  of  Eu- 
rope, the  remainder  is  65  million.  How  many  inhabitants  has 
each  continent  ? 

28.  The  length  of  the  Rio  Grande  River  is  f  that  of  the 
Volga,  and  the  Mississippi  is  600  miles  less  than  twice  as  long 
as  the  Volga.  If  \  the  length  of  the  Mississippi  be  subtracted 
from  that  of  the  Rio  Grande,  the  remainder  is  400  miles. 
Find  the  length  of  each  river. 

29.  In  1900  the  total  wealth  of  the  United  States  was  1532 
million  dollars  more  than  13  times  as  great  as  in  1850,  and 
9016  million  more  than  twice  as  great  as  in  1880.  The  total 
wealth  in  1880  was  174  million  less  than  6  times  as  great  as 
in  1850.     What  was  the  wealth  in  each  of  the  three  years  ? 

30.  The  money  circulation  of  the  United  States  in  1880 
was  13  million  dollars  more  than  60  times  that  in  1800  and 
in  1905  it  was  188  million  more  than  150  times  that  in  1800. 
One-seventh  of  the  amount  in  1880  plus  \  the  amount  in  1900 
was  786  million.     Find  the  circulation  for  each  year. 

31.  The  total  bank  deposits  in  the  United  States  in  1905 
were  3127  million  dollars  less  than  twice  as  great  as  in  1900 
and  681  million  more  than  5  times  as  great  as  in  1880.  The 
deposits  in  1880  were  5105  million  less  than  in  1900.  Find 
the  deposits  for  each  of  the  three  years. 


82  INVOLVED  NUMBER   EXPRESSIONS 

32.  The  amount  of  deposits  in  savings  banks  in  the  United 
States  in  1905  was  703  million  dollars  greater  than  in  1900, 
and  183  million  less  than  4  times  that  in  1880.  The  amount 
in  1900  was  67  million  less  than  3  times  as  great  as  in  1880. 
Find  the  deposits  for  each  of  the  three  years. 

33.  The  total  value  of  the  farms  in  the  United  States  in 
1880  was  280  million  dollars  more  than  3  times  their  value  in 
1850,  and  8333  million  less  than  their  value  in  1900.  The 
value  of  the  farms  in  1880  was  1924  million  less  than  ^  their 
value  in  1900.     Find  the  value  in  each  of  the  three  years. 

34.  The  value  of  the  manufactures  in  the  United  States  in 
1900  was  811  million  dollars  more  than  12  times  that  in  1850, 
and  3071  million  less  than  3  times  that  of  1880;  while  the 
value  in  1880  was  744  million  less  than  6  times  that  in  1850. 
Find  the  value  of  the  manufactures  for  each  year. 

MULTIPLICATION   OF  POLYNOMIALS 
82.    Illustrative    Problem.    A   rectangular   field   is    12   rods 
longer  than  it  is  wide ;  a  second  rectangular  field  which  is  4 
rods  shorter  and  2  rods  wider  than  the  first  has  an  area  of  80 
square  rods  less.     What  are  dimensions  of  the  first  field  ? 

Solution.     Let  w  =  number  of  rods  in  the  width  of  first  field. 

Then        w  +  12  =  number  of  rods  in  the  length  of  first  field, 
and        w  (to  +  12)  =  width  x  length,  or  area  of  first  field ; 
also  w  +  2  =  width  of  second  field, 

and  w  +  8  =  length  of  second  field. 

Hence  (w  +  2)  (w  +  8)  =  width  x  length,  or  area  of  second  field. 

But  the  area  of  the  second  field  is  80  square  rods  less  than  that  of 
the  first. 

Hence  (w  +  2)  (w  +  8)  =  w  (w  +  12)  -  80.  (1) 

To  solve  this  equation  it  is  necessary  to  obtain  the  product 
of  the  two  binominals  w  -+-  2  and  w  +  8  without  first  combining 
the  terms  of  each  binominal.     In  order  to  determine  how  these 


3-5 

3-8 

5-5 
5 

5-8 
8 

MULTIPLICATION  OF  POLYNOMIALS  83 

are  to  be  multiplied,  let  us  consider  two  binomials  in  each  of 
which  the  terms  can  be  combined  if  desired. 

E.g.  Consider  the  product  of  the  binomials  5+3  and  5  +  8.  This 
may  be  represented  by  the  area  of  a 
rectangle  8  feet  wide  and  13  feet 
long.  That  is,  (5  +  3)  (5  +  8)  =  8  .  13 
=  104  square  feet.  But  such  a 
rectangle  may  be  divided  into  four 
rectangles,  as  in  the  figure. 

Hence  the  area  may  be  expressed 
as  the  sum  of  four  areas.  Thus, 
(5  +  3)(5  +  8)  =  5  •  5  +  5  •  8  +  3  •  5  +  3  •  8  =  104,  as  before. 

83.  The  product  5  •  5  is  abbreviated  to  52,  the  small  figure 
indicating  that  5  is  to  be  used  as  a  factor  twice.  It  is  read 
the  square  of  5,  or  5  squared,  since  it  represents  the  area  of  a 
square  whose  sides  are  each  5. 

The  second  method  here  used  for  multiplying  (5  +-3)(5  +8) 
is  applicable  to  any  similar  case,  and  does  not  depend  upon  the 
possibility  of  first  combining  the  terms  of  the  binomials. 

Applying  this  method  to  the  binomials  in  equation  (1)  of  the 
problem  in  §  82,  we  have 

w2  +  8  to  +  2  w  +  16  =  to2  +  12  w  -  80.  (2) 

Subtracting  to2  from  both  sides  and  applying  Principle  I, 

10  w  +  16  =  12  w  -  80.  (3) 

Subtracting  10  to  from  both  sides  and  adding  80  to  both  sides, 

96  =  2  w.  (4) 

Hence  by  D,      48  =  to,  the  width  of  the  field ; 
and  60  =  w  +  12,  the  length  of  the  field. 

Check  by  substituting  to  =  48  in  equation  (1),  and  also  by  showing 
that  the  numbers  48  and  60  satisfy  the  conditions  stated  in  the 
problem. 


84 


INVOLVED  NUMBER  EXPRESSIONS 


84.  In  a  manner  similar  to  that  just  illustrated  we  may  mul- 
tiply two  trinomials. 

E.g.  The  product  of  a  +  b  +  c  and  m  4-  n  +  r,  in  which  the  letters 
represent  any  positive  numbers,  may  represent  the  area  of  a  rectangle, 
divided  into  small  rectangles  as  follows : 


a 

am 

an 

ar 

b 

bm 

bn 

br 

c 

cm 

en 

cr 

Hence,  the  product  is  : 
(a  +  b  +  c)  (m  4-  n  +  ?•)  =  am  +  bm  +  cm  +  an  +  bn  +  en  +  ar  +  br  +  cr, 
in  which  each  term  of  one  trinomial  is  multiplied  by  every  term  of 
the  other,  and  the  products  are  added. 

Evidently  the  same  process  is  applicable  to  the  product  of 
two  such  polynomials  each  containing  any  number  of  terms. 


EXERCISES  AND   PROBLEMS 

Find  each  of  the  following  products  in  two  ways 


1.  (3  +  7  +  10)(2  +  6). 

2.  (5  +  ll)(13  +  10  +  5). 

3.  (6-f-ll)(6  +  7). 

4.  (15  +  8)(15  +  4). 


5.  (7  +  13)(a  +  6). 

6.  (ra  +  ft)(ll  +  5+4). 

7.  42  -36  =  (40  +  2)(30  +  6). 

8.  28- 73  =  (20 +  8)  (70 +  3). 


Find  as  many  as  possible  of  the  following  indicated  products 
in  two  ways : 


9.  (a  +  b)(c  +  d). 

10.  (a;  +  4)(x  +  3). 

11.  (x  +  y  +  z)(a  +  b  +  c). 

12.  (ll  +  13)(r  +  f). 

13.  (5  +  z)(5  +  7). 

14.  (3  +  8)(2  +  4  +  6). 


15.  (aj  +  7)(3a?  +  4). 

16.  (a  +  4)(3a  +  l). 

17.  (3  +  x)(2-\-5x). 

18.  (a  +  6)(3a  +  7&). 

19.  (x  +  y)(2x  +  3y). 

20.  (7x  +  4x)(x  +  8). 


MULTIPLICATION  OF  POLYNOMIALS  85 

21.  A  rectangle  is  7  feet  longer  than  it  is  wide.  If  its 
length   is   increased  by  3   feet   and   its    width  increased   by 

2  feet,  its  area  is  increased  by  60  square  feet.  What  are  its 
dimensions  ? 

22.  A  field  is  10  rods  longer  than  it  is  wide.  If  its  length 
is  increased  by  10  rods  and  its  width  increased  by  5  rods,  the 
area  is  increased  by  630  square  rods.  What  are  the  dimen- 
sions of  the  field  ? 

23.  A  farmer  has  a  plan  for  a  granary  which  is  to  be 
12  feet  longer  than  wide.  He  finds  that  if  the  length  is 
increased  8  feet  and  the  width  increased  2  feet,  the  floor 
space  will  be  increased  by  160  square  feet.  What  are  the 
dimensions  ? 

24.  If  the  length  of  a  rectangular  flower  bed  is  increased 

3  feet  and  its  width  increased  1  foot,  its  area  will  be  increased 
by  19  square  feet.  What  are  its  present  dimensions,  if  its 
length  is  4  feet  greater  than  its  width  ? 

85.  Polynomials  with  Negative  Terms.  The  polynomials  mul- 
tiplied in  the  foregoing  exercises  contain  positive  terms  only. 
The  same  process  is  applicable  to  polynomials  containing  nega- 
tive terms,  as  is  seen  in  the  following  examples : 

Ex.  1.  Find  the  product  of  (7  -  4)  and  (3  +  5).  This 
product,  written  out  term  by  term,  would  give 

(7  +  -4)(3  +  5)  =  7- 3  +  7- 5  + -4- 3  + "4. 5 
=  21  +  35-12-20  =  24. 
Also       (7-4)(3  +  5)  =  3-8  =  24. 

Ex.  2.     Multiply  7  -  4  and  8  -  3. 

(7  _  4)(8  _  3)  =  (7  +  "4)(8  +  ~3)  =  7  •  8  +  7  •  "3  +  "4  •  8  +  "4  .  "3 

=  56  -  21  -  32  + 12  =  15. 
Also  (7-4)(8-3)=3-5  =  15. 


86  INVOLVED  NUMBER  EXPRESSIONS 

EXERCISES 

Find  each  of  the  following  products  in  two  ways,  as  in  the 
above  examples : 

1.  (ll-7)(6  +  5).  5.  (2-3)(46-76). 

2.  (22  -  13) (3  +  7).  6.  (3  +  aj)(7*-3aj). 

3.  (8-5)(7-3).  7.  (8-3)(8-2). 

4.  (17-9)(29-4).  8.  (9-13)(9-17). 

Perform  the  following  indicated  operations : 
9.   (a-6)(c  +  d).  13.   (a  — 6)(7a  +  3  6). 

10.  (a  — 6)(c  — d).  14.  (5  —  y)(5  x  +  3  y). 

11.  (a  -  4) (a  —  5).  15.   (2  a  —  3  b  +  c)(w  +  n). 

12.  (a +  6  —  c)(m  —  n).  16.   (y  —  t)(7v  —  5t). 

86.   The  preceding  exercises  illustrate  the  following  principle : 

Principle  XIII.  The  product  of  two  polynomials  is  found 
by  multiplying  each  term  of  one  by  every  term  of  the  other, 
and  adding  these  products. 

In  case  some  of  the  partial  products  are  negative,  these  are 
combined  by  Principle  IX. 

If  there  are  similar  terms  in  either  polynomial,  these  should 
usually  be  added  first,  thus  putting  each  polynomial  in  as  sim- 
ple form  as  possible. 

E.g.  (3x  +  2  -  2x)(4  x  +  3  -  3x)  =  (x  +  2)(x  +  3) 

=  a;2  +  2z  +  3x  +  6=a;2  +  5ar  +  6. 

EXERCISES  AND   PROBLEMS 

Perform  the  following  indicated  operations : 

1.  (a-7)(3a;-4  +  8). 

2.  (l-2a?  +  x-3){2x  +  ±a  +  lx). 

3.  (4a  —  x  —  3a)(2x  +  4:a  +  7x). 


MULTIPLICATION  OF  POLYNOMIALS  87 

4.  (5x  +  3y-4:X-2y)(6y  +  3x-2y  +  y). 

5.  (13a-6-12o)(2&-3a). 

6.  (xy-5xy  +  4:y)(Sy-3-7y). 

7.  (llao  +  3a)(2&-3&  +  5). 

8.  (6-4aj  +  3*)(7aj  +  y  — 3«  +  l). 

9.  (13a  —  12x-y  +  3)(5x-3y  +  xy  +  5). 

10.  (37  —  13w  +  a)(a  —  w  +  8). 

11.  (x-2+y)(4:y-3x). 

12.  (ll&-a-10&)(6a-3&-2a). 

13.  (7  +  y-x)(2y  +  x-l). 

14.  (5as  +  3y-l)(a;-2). 

15.  (—8a- 1  +  7  a) (5a -8  — 3a). 

Solve  the  following  equations,  in  each  case  verifying  the 
solution  by  substituting  in  the  given  equation  the  result  found 
for  the  unknown  number. 

16.  (a?  +  2)  (a;  +  3)  =  (x  -  3)  (x  + 10)  +  10. 

17.  (5a;-4)(6-aj)-97  =  (a;-l)(6-5a;). 

18.  (3  n  -1)  (18  -  ft)  =  (ft  +  6)  (16  -  3  n). 

19.  (7-a)(9a-8)  =  31  +  (36-9a)(a  +  2). 

20.  (4  a  +  4)  (a  -  3)  =  (4  a  +  1)  (a  +  7)  - 13  a  +  221. 

21.  (n  +  6)(3  n  -  4)  - 14  =  («  +  8)  (3  n  -  3). 

22.  (8  n  +  6)  (10  -  w)  +  150  =  (1  -  w)  (8  n  +  3). 

23.  (a-l)(13-6a)  =  (6a-3)(8-a)-21. 

24.  (7  a;  - 13)(6  -  a;)  -  (a; +  4)(3- 7  a;)  =  70. 

25.  A  club  makes  an  equal  assessment  on  its  members  each 
year  to  raise  a  certain  fixed  sum.  One  year  each  member 
pays  a  number  of  dollars  equal  to  the  number  of  members  of 
the  club  less  175.  The  following  year,  when  the  club  has  50 
more  members,  each  member  pays  $  5  less  than  the  preceding 
year.  What  was  the  membership  of  the  club  the  first  year 
and  how  much  did  each  pay  ? 


88  INVOLVED  NUMBER   EXPRESSIONS 

26.  There  are  two  numbers  whose  difference  is  6  and  whose 
product  is  180  greater  than  the  square  of  the  smaller.  What 
are  the  numbers  ? 

27.  There  are  four  consecutive  even  integers  such  that  the 
product  of  the  first  and  second  is  40  less  than  the  product  of 
the  third  and  fourth.     What  are  the  numbers  ? 

28.  There  are  four  consecutive  integers  such  that  the  prod- 
uct of  the  first  and  third  is  223  less  than  the  product  of  the 
second  and  fourth.     What  are  the  numbers  ? 

29.  There  are  four  numbers  such  that  the  second  is  5 
greater  than  the  first,  the  third  5  greater  than  the  second,  and 
the  fourth  5  greater  than  the  third.  The  product  of  the  first 
and  second  is  250  less  than  the  product  of  the  third  and 
fourth.     What  are  the  numbers  ? 

30.  Prove  that  for  any  four  consecutive  integers  the  prod- 
uct of  the  first  and  fourth  is  2  less  than  the  product  of  the 
second  and  third. 

SQUARES  OF  BINOMIALS 

87.  Just  as  x2  is  written  instead  of  x  •  x,  so  (a  +  6)2  is  written 
instead  of  (a  +  b)  (a  +  6).  The  square  of  a  binomial  is  found 
by  multiplying  the  binomial  by  itself  as  in  §  86. 

E.g.   (a  +  b)2  =  (a  +  b)(a  +  b)  =  a2  +  ab  +  ab  +  b\ 

Hence  (a  +  6)2  =  a2  +  2  ab  +  b2. 

This  product  is  illustrated  in  the  accompanying  figure,  and 
is  evidently  a  special  case  of  the  type  exhibited  in  the  figure, 
page  83. 

Translated  into  words  this  identity  is : 
The  square  of  the  sum  of  any  two  numbers  is 
equal  to  the  square  of  the  first  plus  twice  the 
product  of  the  two  numbers  plus  the  square  of 
the  second. 


ba 

b3 

an- 

ab 

SQUARES   OF  BINOMIALS  89 

88.  Similarly  we  obtain  the  square  of  the  difference  of  two 
numbers.  («_*)»s  «»-*«*  +  *• 

Translate  this  identity  into  words. 

While  these  squares  are  ordinary  products  of  binomials  and  can 
always  be  obtained  according  to  Principle  XIII,  they  are  of  special 
importance  and  should  be  studied  until  they  can  be  reproduced  from 
memory  at  any  time. 

EXERCISES  AND  PROBLEMS 

Perform  the  following  indicated  operations : 

1.  (a  +  6)2.  4.    (a-2)2.  7.  (4  +  9)2. 

2.  (17 -3)2.  5.    (21  -bf.  8.  (c-f)2. 

3.  (6  + a)2.  6.   (z-7)2.  9.  (x-\)\ 

10.  (r  —  s)2  —  (r-f  s)2+(r  —  s)(r  +  s). 

Check  each  of  the  above  by  substituting  special  values  for 
the  letters  and  combining  the  terms  of  the  binomial  before 
squaring. 

11.  Find  the  square  of  42  by  writing  it  as  a  binomial,  40  +  2. 

12.  Square  the  following  numbers  by  writing  each  as  a 
binomial  sum :  51,  53,  93,  91,  102,  202,  301. 

13.  Find  the  square  of  29  by  writing  it  as  a  binomial,  30  —  1. 

14.  Square  the  following  numbers  by  first  writing  each  as  a 
binomial  difference :  28,  38,  89,  77,  99,  198,  499,  998,  999. 

Solve  the  following  equations,  verifying  each  solution : 

15.  (a  +  4)2  +  (a  - 1)  (2  a  +  5)  =  (a  +  4)  (3  a  +  2) . 

16.  (a  -  1)  (3  a  - 1)  -  (a +  1)2  =  2  a2- 18. 

17.  (6  -  a)2  -f  (a  -  3)(2  a  -  5)  =  (3  a  + 1)  (a  -  3)  +  84. 

18.  (7  a  -18)(a  -f  4)  -  (a  - 1)2  =  6  (a  + 1)2  -  79. 


90  INVOLVED  NUMBER  EXPRESSIONS 

19.  (2&-30)(&-l)-5&2  =  6  6-3(7>  +  5)2  +  65. 

20.  (5-&)(6&  +  5)+4(6-3)2  =  20-2(7>  +  l)2  +  3  +  16&. 

21.  (5  -  c)2  +  (7  -  c)2  +  (9  -  c)2=(c  -  1)(3  c-58)  -  93. 

22.  (5  c  -  3)  (2  +  c)  -  4  (c  -  l)2  =  (c  +  1)2+  54. 

23.  (8  -  4  c) (5  -  c)  =  (c  + 1)2  +  (c  +  3) (3  c  -  8)  +  218. 

24.  (4y-9)(y-5)-5(2/-4)2=(8-2/)(4  +  2/)-82. 

25.  (y  +  6)"-3(y-l)»+(4-y)(5--2y)  =  25y  +  9. 

26.  (y-l)2  +  4(i/  +  l)2  +  (l-2/)(5y  +  6)  =  15y-29. 

27.  a(a;  +  3)  +  (a;  +  l)(a;  +  2)  =  2a;(a:-f-5)+2. 

28.  aj2=(*-3)(o:  +  6)-12. 

29.  (5  +  5x)(3-x)+2(x  +  l)2+3(x  +  l)(x-7)  =  17(x+l). 

30.  (8  +  3  *)(4  -  x)  +  (a?  -  l)(aj  -  2)  +  2  (as  +  5)2  =  105. 

31.  There  is  a  square  field  such  that  if  its  dimensions  are 
increased  by  5  rods  its  area  is  increased  625  square  rods.  How 
large  is  the  field  ? 

Suggestion :  If  a  side  of  the  original  field  is  w,  then  its  area  is  to2, 
and  the  area  of  the  enlarged  field  is  (w  +  5)2. 

32.  A  rectangle  is  9  feet  longer  than  it  is  wide.  A  square 
whose  side  is  3  feet  longer  than  the  width  of  the  rectangle  is 
equal  to  the  rectangle  in  area.  What  are  the  dimensions  of 
the  rectangle? 

33.  A  boy  has  a  certain  number  of  pennies  which  he 
attempts  to  arrange  in  a  solid  square.  With  a  certain  num- 
ber on  each  side  of  the  square  he  has  10  left  over.  Making 
each  side  of  the  square  1  larger,  he  lacks  7  of  completing  it. 
How  many  pennies  has  he  ? 

34.  A  room  is  7  feet  longer  than  it  is  wide.  A  square  room 
whose  side  is  3  feet  greater  than  the  width  of  the  first  room 
is  equal  to  it  in  area.  What  are  the  dimensions  of  the  first 
room  ? 


SQUARES  OF  BINOMIALS  91 

35.  Find  two  consecutive  integers  whose  squares  differ 
by  51 

36.  Find  two  consecutive  integers  whose  squares  differ 
by  97. 

37.  Find  two  consecutive  integers  whose  squares  differ  by  a. 
Show  from  the  form  of  the  equation  obtained  that  a  must  be 
an  odd  integer. 

38.  There  are  four  consecutive  integers  such  that  the  sum 
of  the  squares  of  the  last  two  exceeds  the  sum  of  the  squares 
of  the  first  two  by  20.     What  are  the  numbers  ? 

39.  Two  square  pieces  of  land  require  together  360  rods  of 
fence  ?  If  the  difference  in  the  area  of  the  pieces  is  900  square 
rods,  how  large  is  each  piece  ?     (Hint :  x2  —  (90—  a;)2  =  900.) 

40.  There  is  a  square  such  that  if  one  side  is  increased  by 
12  feet  and  the  other  side  decreased  by  8  feet  the  resulting 
rectangle  will  have  the  same  area  as  the  square.  Find  the 
side  of  the  square. 

41.  A  regiment  was  drawn  up  in  a  solid  square.  After  50 
men  had  been  removed  the  officer  attempted  to  draw  up  the 
square  by  putting  one  man  less  on  each  side,  when  he  found 
he  had  9  men  left  over.     How  many  men  in  the  regiment  ? 

42.  There  is  a  rectangle  whose  length  exceeds  its  width 
by  11  rods.  A  square  whose  side  is  5  rods  greater  than  the 
width  of  the  rectangle  is  equal  to  it  in  area.  What  are  the 
dimensions  of  the  rectangle  ? 

89.  Thus  far  the  processes  of  algebra  have  all  been  based 
upon  thirteen  fundamental  principles,  together  with  certain 
obvious  form  changes  indicated  in  §  37.  These  latter  are  of 
general  use  in  elementary  arithmetic  and  need  no  special  em- 
phasis here.  The  principles,  however,  for  the  most  part  refer 
to  methods  not  common  in  arithmetic.  These  should  now  be 
carefully  reviewed  and  a  list  of  them  made  for  convenient 
reference. 


92  INVOLVED  NUMBER   EXPRESSIONS 

REVIEW  QUESTIONS 

1.  How  would  3  •  5  and  7  •  5  be  added  in  arithmetic  ?  Why 
cannot  3n  and  In  be  added  in  the  same  manner?  State  in 
full  the  principle  by  which  3n  and  In  are  added.  In  this 
example  what  number  is  represented  by  n  ?  Test  the  identity 
3n  +  7w=10nby  substituting  any  convenient  value  for  n. 

2.  How  is  5  •  9  subtracted  from  11  •  9  in  arithmetic  ?  In 
what  different  manner  may  this  operation  be  performed? 
Why  is  it  sometimes  necessary  to  perform  subtraction  in  the 
second  way  ?  State  in  full  the  principle  by  which  12  x  is  sub- 
tracted from  31  x.  In  the  identity  31  x  — 12  x  =  19  x,  what 
number  is  represented  by  #?  Test  the  equality  by  substituting 
any  convenient  number  for  x. 

3.  How  is  the  product  2-3-5  multiplied  by  4  in  arithmetic  ? 
In  what  different  way  may  this  multiplication  be  performed  ? 
Why  should  it  ever  be  performed  in  the  second  way  ?  State 
in  full  the  principle  by  which  2  ax  is  multiplied  by  3. 

4.  How  is  11  +  3  multiplied  by  4  in  arithmetic  ?  In  what 
different  way  may  this  operation  be  performed?  Why  is  it 
sometimes  necessary  to  multiply  in  the  second  way  ?  State  in 
full  the  principle  by  which  a  +  8  is  multiplied  by  7. 

What  principles  are  used  in  performing  the  following  indi- 
cated operations  ? 

ax  +  bx  +  cx,  aby+cby,  c(4a  +  2&), 

3y-by  +  cy,  41  a  -32  b-  17  a  +  SOb. 

5(16a-36),  -3(6x-7y),  a  (11 -6). 

5.  Divide  2-4-6-20  by  2  without  first  performing  the 
multiplication  indicated  in  2-4-6-20.  Do  this  in  several 
ways  and  show  that  all  the  quotients  obtained  are  equal. 
State  in  full  the  principle  used. 

What  principles  are  used  in  the  following  operations  ? 
3  •  5  ab  =  15  ab,  c  •  5  b  =  5  (be),  20  ab  +  4  =  5  ab, 

16xy  +  x  =  l&y,        3*  +  15*  =  18<,  78^-41  A  =  37  ft. 


REVIEW    QUESTIONS  93 

6.  How  is  12  + 18  divided  by  6  in  arithmetic  ?  In  what 
different  way  may  this  division  be  performed  ?  Why  is  it 
sometimes  necessary  to  perform  division  in  the  second  way  ? 
State  in  full  the  principle  used  in  performing  the  operation 
(6  x  +  9  y)  -5-  3.  What  principle  is  used  in  performing  each  of 
the  following  indicated  operations  ? 

5(a  +  x),        3y  —  4:y,        (24  x  +  9  y)  -r-  3,        5x+ax. 

7.  Define  equality ;  equation ;  identity.  State  in  detail 
how  the  equation  and  the  identity  differ.  Give  an  example 
of  each. 

8.  In  what  ways  may  an  equation  be  changed  into  another 
equation  such  that  any  number  which  satisfies  either  also 
satisfies  the  other  ? 

Describe  some  of  the  operations  which  change  the  form  of 
the  members  of  an  equation,  but  not  their  value. 
State  Principle  VIII  in  full. 

9.  Name  several  pairs  of  opposite  qualities  all  of  which 
are  conveniently  described  by  the  words  "  positive  "  and  "  nega- 
tive." What  symbols  are  used  to  replace  these  words  when 
applied  to  numbers  ? 

10.  When  loss  is  added  to  profit,  is  the  profit  increased  or 
decreased  ?  What  algebraic  symbols  are  used  to  distinguish 
the  numbers  representing  profit  and  loss  ? 

11.  Give  an  illustration  by  means  of  the  number  scale  to 
show  that  a  number  may  represent  either  a  change  of  position 
in  one  direction  or  the  other,  or  &  fixed  position  with  respect  to 
the  zero  point. 

12.  Why  do  we  call  positive  and  negative  numbers  signed 
numbers  ?    What  is  meant  by  the  absolute  value  of  a  number  ? 

13.  State  Principle  IX  in  full. 

14.  By  means  of  the  number  scale  describe  the  "counting" 
method  of  adding  signed  numbers 


94  INVOLVED  NUMBER  EXPRESSIONS 

15.  Make  a  list  of  pairs  of  opposite  qualities  to  which  posi- 
tive and  negative  numbers  apply.  State  in  each  case  what  is 
represented  by  positive  and  what  by  negative  numbers. 

16.  How  is  the  correctness  of  subtraction  tested  in  arith- 
metic ?  Is  the  same  test  applicable  to  subtraction  in 
algebra  ? 

17.  Explain  subtraction  by  counting  on  the  number  scale. 

18.  How  do  negative  numbers  make  subtraction  possible  in 
cases  where  it  is  impossible  in  arithmetic  ? 

19.  What  is  a  convenient  rule  for  subtracting  signed  num- 
bers ?     State  Principle  X. 

20.  Give  examples  of  equations  which  could  not  be  solved 
without  negative  numbers  and  show  that  such  equations  can 
be  solved  by  means  of  negative  numbers. 

21.  Give  an  example  in  which  positive  and  negative  num- 
bers are  multiplied.     State  Principle  XI. 

22.  Define  division.  How  do  we  obtain  the  law  of  signs  in 
division  ?  State  Principle  XII.  What  is  the  test  of  the  cor- 
rectness of  division  ? 

23.  What  may  be  the  significance  of  a  negative  number 
when  obtained  as  a  result  of  solving  a  problem  ? 

24.  Explain  how  one  set  of  signs  +  and  —  can  be  used  to 
indicate  both  quality  and  operation. 

Show  that  a  -f-  ~b  =  a  —  b  and  a  —  +b  =  a  —  b. 

25.  What  is  a  polynomial  ?  A  term  ?  How  are  poly- 
nomials classified?  What  are  similar  terms?  By  what  prin- 
ciple are  similar  terms  added  ?  By  what  principle  are  they 
subtracted  ?  In  adding  or  subtracting  polynomials,  how  may 
the  terms  be  arranged  for  convenience  ?  What  is  the  prin- 
ciple for  removing  a  parenthesis  when  preceded  by  the  sign  +  ? 
By  the  sign  —  ?     State  Principle  VII  in  full. 


REVIEW  EXERCISES  95 

26.  Make  a  diagram  to  show  how  to  multiply  (7  +  4)  by 
(11  +  8)  without  first  uniting  the  terms  of  the  binomials. 
Multiply  (a +  6)  by  (c  +  cT)  in  the  same  manner.  Multiply 
(12  —  3)  by  (9  —  7)  in  two  ways  and  compare  results.  State 
the  principle  by  which  two  polynomials  are  multiplied. 

27.  Explain  why  x2  is  called  the  square  of  x  or  x  squared. 
■  State  in  words  what  is  the  square  of  the  binomial  (x  +  a)  ;  of 

the  binomial  (x  —  a). 

28.  Show  by  an  example  how  negative  numbers  may  be  used 
in  solving  a  problem,  even  though  the  answer  to  the  problem 
is  positive  and  the  statement  of  the  problem  does  not  involve 
negative  numbers. 

REVIEW  EXERCISES 

In  the  following  exercises  perform  the  indicated  operations, 
remove  all  parentheses,  solve  all  equations,  verify  the  results, 
and  in  each  case  state  the  principles  used. 

1.  Add  3x  +  4y  —  3z,  5x  —  2y  —  z,  and  3y  —  5  a;  +  7  z. 

2.  From  15  a  +  4  6  —  13  be  subtract  3  a  —  8  b  +  2  be. 

3.  Subtract  7  x  —  5y  —  7  a  from  6x  +  5y  +  3a. 

4.  (3x-4:y-z)(x  +  y  +  z).       5.    (6  -5)(2  a2b  -3  ab-  b). 

6.  Add  11  axy  + 13  x  — 14  y,  2  y  —  4  x,  and  3  y  +  x  —  8  axy. 

7.  (5  x  -  3  b)  +  (2  x  +  b)  -  (4  x  -  2  b  -  x  +  5  6). 

8.  Add  19  b  +  3  c,  2  b  -  7  c,  2  c  - 14  b,  and  c  +  8  6. 

9.  -(a  —  3  6—  c)  —  (2  c—  a-  5  6)  +  (a-  c  +  6). 
10.    Subtract  2 a; -f- 4 y  +  z  from  13 a  —  3y  —  5z  +  8. 

"•    3  +  6  +  12  +  T  =  " 

12.  5(^-7)  -3(14-z)  +60=  l-10ar. 

13.  13  (!-»)- 6  (2  a; -5)  =80 +  12  a:. 


96  INVOLVED  NUMBER  EXPRESSIONS 

14     x-3     s-4     a?-8     a?-6  =  18 

7  5  2^6 

15.  Add  7aj-3y-4,   5a;  +  2y-f-5,    and  3y-Sx-6. 

16.  Add  13  a  +  4  b  -  9  c,  2  c  -  8  b  - 16  a,  and  8  a  -  5  6  -  8  c. 

17.  (13-4-2)(5*-3  +  4aj). 

18.  5(x-y)2  -5(y-xf  +  (x-y)(x  +  y). 

19.  Square  73,  79,  92,  98,  1003,  and  995  as  binomials. 

20.  From  17  6 -4  a -2  c  -  19  subtract  8  c-5  a  — 8  6  +  4. 
2i.  3 -(3 -2 +  6 +  8 -3) +  8- (9 -3 +  8). 

22.  3(4-a)-2(5-6a;)  =  8<c  +  4. 

Aft  0*         -      />•  /y»  iM 

23.  _4--4-___4-_  =  —  2. 
4     8      16     2     32 

24     y-3     y  +  9  =  y-ll     2 
*       4  10  4 

25.  ^i  +  ^±J  +  ^+8  =  2y-20. 

3  3  3  * 

26.  5a;-(3a;-2  +  2i/4-a;)  +  13y-(6-3a;  +  4). 

27.  From  3  —  4a  —  5c  +  8ar*  subtract  2a?2—  2a  —  4c  +  8. 

28.  12  +  (2o-3c-46)-(36-c-a-8). 

29.  y+y±^+y±5=25. 

30.  r-fciJ!+fc=»+«hI8.iBL 

3  5  5  2 

31.  Add  y  -  20,  4  y  +  6,  2  y  +  4  x  - 13,  and  2  a  -  8  y  -  40. 

32.  (x  - 1)(2  *  -  2)  +  (a?  -  5)2  =  (3  -  a) (24  -  3  a)  -  7. 

33.  Subtract  16  —  a;  +  2z  —  4y  from  3x  —  5z  —  8y. 

34.  19  +  (2  a;  -  7)  -  (31  -  4  a?  -  8  -  2  a?)  =  5  x  +  7. 

35.  (4  aft  —  6ac  —  5  ad)(b  —  c  +  a"). 

36.  (17  x  +  8)(*  -  1)  +  8  =  (2  -  x)(6  -  17  aj)  + 19. 


REVIEW  EXERCISES  97 

37.    5  -  (a  +  b  -  c  -  d  +  8)  +  (3  +  a  +  c  -  d)  -  5. 

38. h- — ! =  a  —  ^o. 

10         10  10 

39.  Add  6  a  +  9,  8  a  - 13,  46  a  -  8,  and  6  -  54  a. 

40.  (a-2)(6a-4)+2(a-l)2=(6-a)(30-8a)  +  4. 

41.  From  6  (a+  2)  +  3(c  +  4)  -  2  (6  -  d)  subtract  2  (a  +  2) 
-2(c  +  4)+3(6-d). 

a  ,  a  +  7     a  —  3  _  a  +  227     1 
42'    3  +  _4  3~~^  1' 

43.  (a-2-3c-8  +  2  6)(6-a-c-&  +  8). 

44.  ^ll  +  ^±l  +  lip3=2 

2  2  12 

45.  a2&  -  (3  6  -  8  a2  -  7)  +  3  a&2  -  (4  ab2  +  8  -  2  a2). 

46.  ^+l  +  ^Z^  +  ^lI  =  2a-26. 

4  4  4 

47.  Add  12  a262c  +  8  ax,  6  ax -8  a2b2c,  and  2  ax  +  3  a262c. 

48.  Add  5  a#2  +  3  a?y  -\-  <ixy,  2  x2y  —  6  xy2  —  3  ay,  and  4  xy. 

49.  Add  6ab  —  3c—  2  a,   2c  —  4  a&  —  5  a,  5c—  a  +  a&,  and 
3  +  5a-2c-3a6. 

CA     fl  +  l.«  +  3.»-l     w  +  13  ,n-2 

5U.     — _ 1 . 1 . —  — _ [_ 


3  4  4  3  3 

51.  (n-4)(6-3n)-(6-w)2-10  =  -4n(n-4). 

52.  From  35  ab  -  8  x  —  9  z  + 13  subtract  16a6  -  4z  +  5  x  +  8. 

53.  (n  +  2)2  +  (n-l)2+(n  +  l)2  =  3n(?i  +  2)  +  60w  +  130. 

54.  Subtract  5a  —  8x—  6y  from  13 x  + 14 y  — 15 z  —  4 a. 

55.  From   9y  —  4a:  —  6  z  —  3  &    subtract  8—  9  y  —  3#  —  2 z. 

56.  2  <e  +  4  -  6  (5  x-  8  -  7  a)  +2  -  4  a  =  6  (2  -3  a;)  -42. 

57.  _  (7  +  4a;-8-2a;)+4-2a;  =  6a;  +  25. 


98  INVOLVED  NUMBER   EXPRESSIONS 

58.  16  +  5x-(8x  +  9-4x  +  17)=8x-3. 

59.  6x-3-(4x+8-9x)-(5x-2)=x  +  ll. 

60.  (a  — 1  +  6  —  c  —  cf)(4a  +  5&  +  3c  —  2d). 

61.  (4ax  — 3a?/-f  5az  — 8)(x  +  ?/  — z  +  2). 

62.  (3ac-2ad  +  4ed)(2  +  3  +  4  +  5). 

63.  7-(3a-2&-4a)+6  +  2a-(3&-2a-a). 

64.  8x  +  (5y-8)  +  (2y-l)-(13y  +  8x-17). 

65.  16  ax  +  4  -  (8  -  8  ax  -  a)  -  (12  ax  -  13  -  ax). 

66.  Add   15  ax2 +  3  6c2,   2  6c2 - 7  ax2,    and   5 +  2  ox2- 5  6c2. 

67.  Addl6-7a6-2a2  +  5a&,  4a2-2a6,  and5a&-8. 

68.  Add  51  x2y  -  35  +  12  a2,   41  -  17  a2  -  57  x*y,  and  3x*y. 

69.  Add  35  b2  -  13  c2,   8  c2  -  3  b2c2,  and   6  &2  -  8  c2  -  9  62c2. 

70.  Add  19-2x  +  3x26  +  26,    4  x2&  +  x  +  5  b  +  8,  and  4x. 

71.  Subtract  2  a  -  6  tfb-3x+2l  from  19  -  2  *  +  3  x2b  -  7  a. 

72.  From  6  a  -  45  -f  8  b  -  3  c  +  82  c&  subtract  7  &  -f  18  +  6  c. 

73.  (17  +  2a-3&-4&c)(2-a  +  &-c). 

74.  (13c-4d  +  8e-3)(c-d). 

75.  (4xy-2y-3x-2)(y-x  +  y  +  5). 

76.  (9  ax  —  3  op  -  5  a  —  2  x  +  4)  (5  —  x). 

77.  From  41  a2  +  7  a&  -  5  c2-  9  ab  subtract  8  ab  +  7  c2+  50  a2. 

78.  From    15  +  2x  —  9xy  —  3y   subtract  5  xy  —  4  a;  —  2 ?/. 

79.  Subtract  12  +  8  x  —  4  a  —6  c  — 18  abc  from  5  x— 2  a  +  3  c 

80.  2-(71z  +  42y-15x-64)-(5-91x-2y-13xy). 

81.  (31-2*  +  3y-5)(aj  +  y). 

82.  8x  +  (13  +  18x-6)-(5-6x)  =  16x+10. 


REVIEW  EXERCISES  99 

83.  (9z-3)(4-aO+(z-3)2=  -8  (z  +  2)2  +94. 

84.  (x  +  l)2  +  (a;  +  2)2  +  (a;  +  3)2=(3a;-l)(a;+12)-43. 

85.  (2  x  +  5) (a;  -7)-(x- 1)2  =  (*  +  l)(x  +  2)  -  28. 

86.  3  (o-xf-  (2x  -  l)(a?  -  !)=(*-  7)  (a;  + 10)  + 17  *  +  50. 

87.  5  -  (7  -  4  x  +  2  ^  -  4  b)  -  (8  a?  -  6  y  -  9  +  2  6)  +  8  a. 

88.  8a>-(-3-2-4-7)  +  5a;  +  (2  +  6  +  4)-(-3a;  +  2). 

89.  5y  +  2a?-(6-4aj-5a;)-3y-(4a;-2y)-(-7y+8). 

90.  35  y  -  (41  x  - 16  - 12  y)  +  5  a;  +  ( -  6  +  46  y  - 18  a?) . 

91.  (.T-l)2-(a?-8)(2a;-l)=-a;2  +  98. 

92.  (32  +  a)(4a-l)  +  (5-a;)2  +  (a;-l)2  =  6(a:  +  l)2  +  194. 

93.  (2a?-7)(5-aj)-(2-5a;)(l  -  a;)  =  -  a;  (7  a?  -  34)  - 17. 

94.  (7  +  x)(x-4)  +  (l-x)2=-23  +  2x2. 

95.  (12-4a)(2-a;)-4(l  +  :z)2  =  5a;  +  119. 

96.  (x  - 17)(59  -  2  x)  -  (1  -  a;)2  =  (6  -  3a;)  (*  -  2)  +  384. 

97.  (3  x-  2)  +  (x - 1)2  +  (x  -  2)2  =  2  (a; -  l)(x -  2)  +  5. 

98.  (6-3  a>)(2  +  x)  + 16  (x  - 1)2  =  13  (x  +  4)2  +  364. 

..  x+8     x- 9  .  x- 17     4a-7  ,  2a;  +  6  .  5-31a; 


12  6  2  3  12 

100.    3j^_3jE±3     a^=aL±_5  20. 

6  3  2  6  3 


CHAPTER   IV 
THE  SOLUTION  OF  PROBLEMS 

90.  Some  of  the  advantages  of  algebra  over  arithmetic  in 
solving  problems  have  been  pointed  out  in  the  preceding  chap- 
ters. For  instance,  brevity  and  simplicity  of  statement  secured 
through  the  use  of  letters  to  represent  numbers ;  the  transla- 
tion of  the  words  and  sentences  of  problems  into  number  ex- 
pressions and  equations  ;  and  the  clear  and  logical  solution  of 
the  equation,  step  by  step. 

91.  A  still  greater  advantage  is  set  forth  in  the  present 
chapter ;  namely,  the  opportunity  offered  by  the  symbols  and 
processes  of  algebra  to  summarize  a  whole  class  of  problems 
under  one  solution,  called  the  formula,  which  is  thereafter  used 
to  solve  all  problems  of  the  class. 

PROBLEMS    INVOLVING   INTEREST 

92.  A  class  of  problems  already  within  the  pupil's  experi- 
ence will  illustrate  this  point.  The  different  cases  of  percentage 
or  interest  have  been  studied  in  arithmetic,  and  a  large  number 
of  isolated  problems  have  been  solved  according  to  the  rules. 
In  this  instance,  therefore,  we  proceed  at  once  to  summarize 
all  of  this  work  in  a  few  short  statements. 

T,etp  =  any  principal,  i.e.  a  number  of  dollars  at  interest. 
i  =  the  interest,  i.e.  the  number  of  dollars  accrued. 
r  =  the  rate,  to  be  expressed  in  hundredths. 
t  =  the  time,  to  be  expressed  in  years  and  fractions  of  a 
year. 

100 


PROBLEMS   ON  INTEREST  101 

Then  the  rule  of  arithmetic  for  finding  the  interest  when  the 
principal,  rate,  and  time  are  given  is 

interest  =  principal  x  rate  x  time, 

i.e.  i=prt.  (1) 

If  in  this  equation  i  =  150,  r  =  .05,  t  =  6,  find  p.  If  i=  190.5,  />  =  635, 
r  =  .03,  find  t.  If  i  =  665,  p  =  1000,  t  =  17,  find  r.  Substitute  other 
values  for  any  three  of  these  letters  and  find  the  value  of  the  remain- 
ing letter. 

Solve  equation  (1)  for  t  in  terms  of  i,  p,  and  r ;  also  for  p  in 
terms  of  i,  r,  and  t,  and  for  r  in  terms  of  p,  t,  and  i. 

It  follows  that  if  any  three  of  the  four  numbers,  p,  r,  i,  and 
t,  are  given,  the  remaining  one  may  be  found. 

Let  the  student  state  four  rules  of  interest  by  translating 
these  formulas  into  words. 

Note  the  simplicity  of  these  equations  compared  with  the  corre- 
sponding rules  in  arithmetic. 

Solve  each  of  the  following  problems  by  substituting  in  the 
proper  formula : 

1.  What  is  the  simple  interest  at  5  %  on  $400  for  5  years 
and  9  months  ? 

Solution.     i=p-r-t  =  400  •  T^  •  5|  =  4  •  5  •  \s  =  115. 

2.  In  what  time  will  the  simple  interest  on  $750  at  3  % 

amount  to  $225?     Substitute  in  the  formula  t  =  — 

pr 

3.  What  is  the  semi-yearly  income  from  an  endowment  of 
$2,700,000,  the  rate  being  4|  of0  per  annum  ?     (Here  t  =  \.) 

4.  A  father  invested  $1500  at  5^%,  the  simple  interest  on 
which  was  to  go  to  his  eldest  son  on  his  21st  birthday.  The 
young  man  received  $1240.  How  old  was  the  son  when  the 
investment  was  made  ? 


102  THE  SOLUTION  OF  PROBLEMS 

5.  What  is  the  amount  of  money  invested  if  it  yields 
$  787.50  interest  per  annum,  the  rate  being  5\%?    (Here  t  =  1.) 

6.  A  certain  investment  yields  $8160  in  8  years.  What  is 
the  principal,  if  the  rate  of  interest  is  4  °fo  ? 

7.  A  $45,800  investment  yielded  $13,396.50  interest  (sim- 
ple) in  6\-  years.     What  was  the  rate  of  interest  ? 

8.  The  endowment  of  a  small  college  is  $  750,000,  the  yearly 
income  from  which  is  $45,000.  What  is  the  average  rate  at 
which  the  endowment  is  invested  ? 

9.  A  capitalist  has  investments  amounting  to  $  360,000,  the 
total  income  from  which  amounts  to  $  1800  per  month.  What 
is  the  average  rate  at  which  the  money  is  invested  ?    (t  =  ^.) 

10.  If  $  700  is  invested  at  5±-  %  simple  interest,  what  is  the 
amount  at  the  end  of  5  years  6  months  ?  This  problem  calls 
for  the  amount,  which  is  the  sum  of  principal  and  interest. 

If  a  =  amount,  then  a=  p  +i  =  p  +  prt. 

11.  Solve  the  equation  a—p+prt  for  p  in  terms  of  a,  r, 
and  t,  and  translate  the  result  into  words. 

12.  Solve  a  =  p  +prt  for  r  in  terms  of  a,  p,  and  t,  and  trans- 
late the  result  into  words. 

13.  Solve  a  =p  +prt  for  t  in  terms  of  a,  p,  and  r,  and  trans- 
late the  result  into  words. 

14.  Seven  years  ago  I  invested  a  certain  sum  of  money  at  6  % 
simple  interest.  The  amount  at  present  is  $5680.  How  much 
did  I  invest  ?      (Use  the  formula  obtained  in  Problem  11.) 

15.  Six  years  ago  A  invested  $470  at  simple  interest.  The 
amount  at  present  is  $  611.     What  is  the  rate  per  cent  ? 

16.  Some  years  ago  I  invested  $500  at  7  %  simple  interest, 
which  now  amounts  to  $815.  How  many  years  ago  was  the 
investment  made  ? 

17.  A  merchant  bought  goods  for  $  250,  and  some  months 
later  sold  the  goods  for  $  300,  making  a  profit  of  1  %  per 
month.    How  many  months  between  the  purchase  and  the  sale  ? 


PROBLEMS   ON  INTEREST  103 

18.  A  real  estate  dealer  sold  a  house  and  lot  for  $  7500, 
for  which  he  received  a  commission  of  4%.  What  was  the 
profit  ? 

In  this  case  the  element  of  time  does  not  enter.  The  word 
commission  is  then  used  for  interest  and  the  principal  is  called 
the  base.     Letting  c  =  commission,  b  =  base,  and  r  =  rate,  we 


have 


c  =  br. 


Applying  this  formula,  c=br  =  7500  •  y^  =  75  •  4  =  300. 

19.  Solve  the  equation  c  =  br  for  b  in  terms  of  c  and  r,  and 
translate  the  result  into  words. 

20.  Solve  c  =  br  for  r  in  terms  of  c  and  b,  and  translate  the 
result  into  words. 

21.  A  merchant's  commission  for  selling  a  carload  of  peaches 
was  $  18.75.  What  was  the  rate  of  commission  if  the  peaches 
brought  $  375  ?     (Use  the  formula  found  in  19.) 

22.  A  broker  received  $25.50  commission  for  selling  $850 
worth  of  bonds.     What  was  his  rate  of  commission  ? 

23.  An  agent  received  3  %  commission  for  buying  and  3|  % 
for  selling  some  property.  He  paid  $  5750  for  it  and  sold  it 
for  $  7200.     What  was  his  total  commission  ? 

24.  How  much  must  I  remit  to  my  broker  in  order  that  he 
may  buy  $  600  worth  of  bonds  for  me  and  reserve  5  °J0  for  his 
commission  ? 

I  must  send  him  both  base  and  commission.  Calling  this  the 
amount  and  representing  it  by  a,  we  have 

a  =  b  +  c  =  b  +  br  =  b  (1  +  r). 
Hence  a  =  600  +  600  .  fa  =  600  +  6  •  5  =  630. 

25.  Solve  the  equation  a=b  +  br  for  b  and  translate  the 
result  into  words. 

26.  Solve  a  =  b  +  br  for  r  and  translate  the  result  into 
words. 


104  THE  SOLUTION  OF  PROBLEMS 

27.  A  merchant  received  $  918  with  which  to  buy  corn  after 
deducting  his  commission  of  2  °J0  on  the  price  of  corn.  How 
much  was  his  commission  and  how  much  was  used  to  buy 
corn  ? 

Here  a  =  980.  Find  b  by  the  formula  under  25,  which  gives  the 
sum  paid  for  corn. 

28.  A  broker  received  $790,  of  which  he  invested  $750 
in  stocks,  reserving  the  balance  as  his  commission.  Find 
the  rate  of  his  commission  by  means  of  the  formula  obtained 
in  26. 

29.  An  agent  received  $  945  with  which  to  buy  lumber  after 
deducting  his  commission  of  5  %  on  the  cost  of  the  lumber. 
How  much  was  his  commission  ?  (First  find  the  base  by  sub- 
stituting in  the  formula  of  25.) 

30.  A  dealer  sold  berries  for  $  18.95,  and  after  deducting  a 
commission  of  2  %  sent  the  balance  to  the  truck  gardener. 
How  much  did  he  remit  ? 

The  sum  he  sent  was  the  difference  between  the  base  and  the  com- 
mission ;  calling  this  d,  we  have 

d=b-c  =  b-br  =  b(l-r). 

Hence  in  this  case  d  =  18.95  (1  —  Tfo)  =  18.57. 

31.  Solve  the  equation  d  =  6(1  —  r)  for  b  in  terms  of  d  and  r 
and  translate  the  result  into  words. 

32.  Solve  the  equation  d  =  b  —  br  for  r  in  terms  of  b  and  d 
and  translate  the  result  into  words. 

33.  After  deducting  a  commission  of  3  %  for  selling  bonds, 
a  broker  forwarded  $  824.50.  What  was  the  selling  price 
of  the  bonds  ?  (Solve  by  means  of  the  formula  under 
Problem  31.) 

34.  A  broker  sold  stocks  for  $  1728  and  remitted  $  1693.44 
to  his  principal.  What  was  the  rate  of  his  commission  ? 
(Solve  by  means  of  the  formula  under  Problem  32.) 


PROBLEMS   ON  AREAS  105 

35.  In  how  many  years  will  $200  double  itself  at  5  %  ? 
In  this  case  i  =p.     Hence,  using  formula  (1),  page  101,  we  have 

200  =  200  x  r§5  x  t. 
Hence  t  =  20. 

36.  In  how  many  years  will  any  sum,  p,  double  itself  at  any 
rate,  r? 

Here  p  =  prt.     Hence,  solving,  t  =  1  +  r. 

37.  In  what  time  will  a  sum  of  money  double  itself  at  6%  ? 
7%?     4i%?     3|%? 

38.  Collect  all  the  formulas  worked  out  in  this  set  of 
problems.  Translate  each  into  words.  Which  were  used 
as  the  original  ones  from  which  the  others  were  deduced  ? 
How  many  and  which  ones  are  needed  in  order  to  derive 
all  the  others  ?  Why  are  such  formulas  better  than  rules 
expressed  in  words  ? 

PROBLEMS  INVOLVING  AREAS 

93.  Another  class  of  problems  already  well  known  to  the 
pupil  in  arithmetic  concerns  the  areas  of  rectangles  and  tri- 
angles. 

If  in  a  rectangle  we  let  the  number  of  units  of  length  be  de- 
noted by  I,  the  number  of  units  of  width  by  w,  and  the  number 
of  square  units  in  the  area  by  a,  then  area  =  length  x  width; 

i.e.  a  =  Iw.  (1) 

If  in  equation  (1)  a  =  144,  I  =  16,  find  w.  If  a  =  1116,  w  =  31,  find  I. 
Substitute  other  values  for  any  two  of  these  letters  and  find  the  value 
of  the  remaining  one. 

Solve  this  equation  for  to  in  terms  of  I  and  a,  and  also  for  I  in 
terms  of  w  and  a. 


106  THE  SOLUTION  OF  PROBLEMS 

Also  if  b  is  the  base  of  a  triangle,  h  its  altitude  (height),  and 
a  its  area,  then  area  =  \  (base  x  altitude) ; 

i.e.        .  .=**.  (2) 

Substitute  particular  values  for  any  two  of  these  letters  and  find 
the  value  of  the  remaining  one. 

Solve  (2)  for  h  in  terms  of  a  and  b,  and  also  for  b  in  terms 
of  a  and  h. 

Let  the  student  translate  each  of  these  equations  into  words. 
Use  these  formulas  in  the  solution  of  the  following  problems : 

1.  How  many  tiles  each  3  by  4  inches  are  needed  to  cover 
a  rectangular  floor  18  by  22  feet  ?     Use  formula  (1). 

2.  How  long  a  space  25  feet  in  width  can  be  covered  by 
340  square  feet  of  roofing  ? 

3.  The  base  and  altitude  of  a  triangle  are  8  and  6  respect- 
ively.    Find  its  area. 

4.  The  base  of  a  triangle  is  12  and  its  area  72.  Find  its 
altitude. 

5.  The  altitude  of  a  triangle  is  16  and  its  area  144.  Find 
its  base. 

6.  A  rectangle  is  5  feet  longer  than  it  is  wide.  If  it  were 
3  feet  wider  and  2  feet  shorter,  it  would  contain  15  square  feet 
more.     Find  the  dimensions  of  the  rectangle. 

Let  w  equal  the  width;  then  construct  number  expressions  for  the 
length,  width,  and  area  under  the  supposed  conditions. 

7.  A  rectangle  is  6  feet  longer  and  4  feet  narrower  than  a 
square  of  equal  area.  Find  the  side  of  the  square  and  the 
sides  of  the  rectangle. 

8.  The  base  of  a  triangle  is  8  inches  greater  than  its  alti- 
tude. If  the  base  is  increased  by  4  inches  and  the  altitude 
decreased  by  2  inches,  the  area  remains  unchanged.  Find  the 
base  and  altitude  of  the  triangle. 


PROBLEMS   ON  AREAS  107 

9.  A  rectangle  is  14  inches  longer  than  it  is  wide.  If  the 
width  is  increased  by  5  inches  and  the  length  decreased  by  4 
inches,  the  area  is  increased  by  70  square  inches.  Find  the 
dimensions  of  the  rectangle. 

10.  A  rectangle  is  15  rods  longer  and  10  rods  narrower  than  an 
equivalent  square.    What  are  the  dimensions  of  the  rectangle? 

11.  The  altitude  of  a  triangle  is  16  inches  less  than  the  base. 
If  the  altitude  is  increased  by  3  inches  and  the  base  by  2  inches, 
the  area  is  increased  by  52  square  inches.  Find  the  base  and 
altitude  of  the  triangle. 

12.  The  width  of  a  rectangular  field  is  20  rods  less  than  its 
length.  If  each  side  is  decreased  by  10  rods,  the  area  is  de- 
creased by  900  square  rods.  What  are  the  dimensions  of  the 
field  ? 

13.  A  picture  is  4  inches  longer  than  it  is  wide.  Another 
picture,  which  is  12  inches  longer  and  6  inches  narrower,  con- 
tains the  same  number  of  square  inches.  Find  the  dimensions 
of  the  pictures. 

14.  A  picture,  not  including  the  frame,  is  8  inches  longer 
than  it  is  wide.     The  area  of  the  frame, 
which  is  2  inches  wide,  is  176  square  inches. 
Find  the  dimensions  of  the  picture. 

15.  A  picture,  including  the  frame,  is 
10  inches  longer  than  it  is  wide.  The  area 
of  the  frame,  which  is  3  inches  wide,  is 
192  square  inches.     What  are  the  dimensions  of  the  picture  ? 

16.  The  base  of  a  triangle  is  11  inches  greater  than  its  alti- 
tude. If  the  altitude  and  base  are  both  decreased  7  inches,  the 
area  is  decreased  119  square  inches.  Find  the  base  and  alti- 
tude of  the  triangle. 

17.  The  base  of  a  triangle  is  3  inches  less  than  its  altitude. 
If  the  altitude  and  base  are  both  increased  by  5  inches,  the 
area  is  increased  by  155  square  inches.  Find  the  base  and  alti- 
tude of  the  triangle. 


— * — ' — ■ — — - — —. 

■!.■ 


x  +  8 

I  Vol/,',/!,,/,  ■/,,„,/,/,/,/,  w,i„/>r/,W, 

'■■..,■:,-) _^__/ Uiu. 


108  THE  SOLUTION  OF  PROBLEMS 

18.  A  commander  in  attempting  to  draw  up  his  men  in  the 
form  of  a  solid  square  finds  that  there  are  80  men  more  than 
enough  to  complete  the  square.  If  he  places  2  more  men  on 
each  side  of  the  square  he  needs  84  more  men  to  complete 
it.     How  many  men  in  his  command  ? 

PROBLEMS  INVOLVING  VOLUMES 

94.  Still  another  class  of  problems  for  which  the  funda- 
mental formulas  are  already  well  known  concerns  the  volumes 
of  rectangular  solids  and  of  pyramids. 

If  the  number  of  units  of  length,  width,  and  height  of  a 
rectangular  solid  be  denoted  by  I,  w,  and  h  respectively,  and  the 
number  of  units  of  volume  by  v,  then 

volume  =  length  x  width  x  height ; 

i.e.  v  =  lwh.  (1) 

If  w  is  the  width,  I  the  length  of  the  rectangular  base  of  a 
pyramid,  h  its  altitude,  and  v  its  volume,  then 

volume  =  ^  (area  of  base  x  altitude)  ; 

Iwh  /ox 

i.e.  "  =  ~3~'  () 

In  equations  (1)  and  (2)  substitute  particular  values  for  any  three 
of  the  letters  and  find  the  value  of  the  remaining  one. 

Use  these  formulas  in  solving  the  following  problems : 

1.  Solve  equations  (1)  and  (2)  for  I,  w,  and  h  respectively 
and  translate  each  equation  into  words. 

2.  How  many  cubic  feet  of  earth  are  removed  in  digging 
a  cellar  18  feet  long,  12  feet  wide,  and  9  feet  deep  ?  Solve  by 
substituting  in  formula  (1). 

3.  If  a  cut  in  an  embankment  is  500  yards  long  and  4  yards 
deep,  how  wide  is  it  if  18,760  cubic  yards  are  removed  ? 


PBOBLEMS   ON   VOLUMES  109 

4.  How  deep  is  a  rectangular  cistern  which  holds  500  cubic 
feet  of  water  if  it  is  6  feet  wide  and  8  feet  long  ? 

5.  The  base  of  a  pyramid  is  16  inches  long  and  12  inches 
wide.     Its  altitude  is  30  inches.     Find  its  volume. 

6.  A  pyramid  whose  volume  is  72  cubic  feet  has  a  base 
whose  area  is  24  square  feet.     Find  the  altitude. 

7.  A  pyramid  whose  volume  is  91  cubic  inches  has  an 
altitude  of  21  inches.     Find  the  area  of  its  base. 

8.  A  rectangular  room  which  is  10  feet  high  is  4  feet 
longer  than  it  is  wide.  If  it  were  5  feet  longer  and  2  feet 
wider,  it  would  contain  950  cubic  feet  more  than  it  does. 
Find  its  length  and  width. 

9.  A  city  building  50  feet  high  extends  back  25  feet 
more  than  its  frontage.  If  the  building  were  8  feet  wider 
and  extended  10  feet  farther  back,  its  capacity  would  be  in- 
creased 59,000  cubic  feet.  What  are  the  ground  dimensions 
of  the  building  ? 

10.  A  pyramid  whose  altitude  is  12  inches  has  a  rectangu- 
lar base  5  inches  longer  than  it  is  wide.  If  the  length  of  the 
base  is  decreased  by  1  inch  and 

the  width  increased  by  2  inches,  /\K 

the  volume  of  the  pyramid  is  in-  //'(,\ 

creased  by  72  cubic  inches.     Find  /  1 I  \  \ 

the  dimensions  of  the  base  of  the  /    /   I    »     \ 

pyramid.  /      /    I    \       \ 

11.  The  base  of  a  rectangular  /        I      j      \  \ 
pyramid    whose    altitude    is    15  \*    /               *^~s     \ 

inches  is   12  inches   longer  than  \/ x+5 ^-\ 

it   is    wide.      If   the   length   and 

width  of  the  base  are  both  decreased  by  3  inches,  the  volume 
is  decreased  by  675  cubic  inches.  Find  the  dimensions  of  the 
base  of  the  pyramid. 


110  THE  SOLUTION   OF  PROBLEMS 

12.  The  base  of  a  rectangular  pyramid  is  15  inches  wide. 
The  altitude  of  the  pyramid  is  4  inches  less  than  the  length 
of  the  base.  If  the  altitude  is  increased  by  6  inches  and  the 
length  of  the  base  by  3  inches  the  volume  is  increased  by  1155 
cubic  inches.  Find  the  altitude  of  the  pyramid  and  the 
length  of  its  base. 

13.  The  altitude  of  a  pyramid  is  7  inches  less  than  the 
length  of  the  base.  The  width  of  the  base  is  13  inches.  If 
the  altitude  and  the  length  of  the  base  are  both  decreased  by 
3  inches,  the  volume  is  decreased  by  364  cubic  inches.  Find 
the  altitude  of  the  pyramid  and  the  length  of  its  base. 

14.  The  width  of  the  base  of  a  pyramid  is  2  inches  greater 
than  its  altitude.  The  length  of  the  base  is  34  inches.  If 
the  altitude  and  the  width  of  the  base  are  both  increased  by 
6  inches,  the  volume  is  increased  by  2992  cubic  inches.  Find 
the  altitude  and  the  width  of  the  base. 


PROBLEMS  INVOLVING  SIMPLE   NUMBER  RELATIONS 

95.  In  the  problems  thus  far  considered  in  this  chapter 
the  formulas  are  already  known  from  previous  study  or  ex- 
perience. Algebra  affords  a  means  of  deriving  such  formulas 
in  many  cases  where  they  are  not  already  known. 

1.  The  sum  of  two  numbers  is  35  and  their  difference  is  5. 
AY  hat  are  the  numbers  ? 

Let  g  =  the  greater  number,  then  35  —  g  =  the  lesser. 

2.  The  sum  of  two  numbers  is  48  and  their  difference  is 
24.     What  are  the  numbers  ? 

3.  The  sum  of  two  numbers  is  41^  and  their  difference  is 
23^.     What  are  the  numbers  ? 

4.  The  sum  of  two  numbers  is  8590  and  their  difference 
is  3480.     What  are  the  numbers  ? 


PROBLEMS   ON  NUMBER   RELATIONS  111 

The  four  problems  just  preceding  are  similar  in  character. 
A  simple  rule  can  be  found  for  solving  all  problems  of  this 
kind.     Consider  problem  1. 

We  have  g  —  (35  —  g)  =  5.  (1) 

By  VII,  2-35  +  g  =  5.  (2) 

By  I,  A,  2  g  =  35  +  5.  (3) 

By  D,  VI,  *=?  +  f'  (4) 

By  F,  g  =  20,  the  greater  number,  (5) 

and  I  =  35  —  g  =  15,  the  lesser  number.        (6) 

The  results  in  the  final  form  tell  nothing  more  than  the 

answers  to  this  particular  problem ;  but  the  value  of  g  in  the 

35      5 
form  g  =  —  4-  -,  if  examined  closely,  tells  much  more.     Stated 

in  full  it  means : 

the  sum  of  the  numbers  .  their  difference 
the  greater  = h ^ 

Examine  now  the  solution  of  the  other  three  problems  to  see 
whether  this  same  expression  will  give  the  greater  number 
in  each  case. 

The  great  advantage  in  not  adding  35  and  5  before  dividing 

35     5 

by  2,  is  that  the  expression 1--  preserves  the  original  num- 
bers as  given  in  the  problem,  so  that  we  see  how  each  enters 
into  the  result. 

If  the  sum  of  the  two  numbers  is  called  s  and  their  differ- 
ence d,  then  we  are  compelled  to  keep  these  letters  separate 
to  the  end  of  the  solution. 

Thus,  the  lesser  is  1  =  s  —  g, 

and  g-(s-g)=d. 

By  VII,  g-s  +  g  =  d. 

By  I,  A,  2g  =  s  +  d. 


112  THE  SOLUTION  OF  PROBLEMS 


s      d 
Hence  by  D,  VI,  the  greater  is  g  =  -  +  -  ,  and  the  lesser  is 


l  =  S-g  =  S-(j  +  f) 


s     d  _  2  s  _  s_d_  s  _d 
2~2~T      2     2~2     2* 


Hence  J  =|  +  g,  am/  /  =  |  -  J- 

These  results  put  into  words  are  as  follows : 

77*e  greater  of  any  two  numbers  is  half  their  sum  plus  half 
their  difference,  and  the  lesser  is  half  their  sum  minus  half  their 
difference. 

Any  problem  of  this  kind  is  solved  by  substituting  in  this 
formula  the  particular  values  given  to  s  and  d  and  simplifying 
the  results. 

E.g.  In  problem  4,  page  110,  s  =  8590  and  d  =  3480.  Hence  the 
greater  number  is  §590  +  §M0=  4295  +  1740  =  6035,  and  the  lesser 

number  is  —  -  —  =  4295  -  1740  =  2555. 
2  2 

5.  Find  by  this  formula  two  numbers  whose  sum  is  17,540 
and  whose  difference  is  11,240. 

6.  Find  two  numbers  whose  sum  is  40  and  whose  difference 
is  52.     Solve  also  without  the  formula. 

Evidently  one  of  these  must  be  a  negative  number  in  order  to 
make  the  difference  more  than  the  sum,  but  the  formula  applies  even 
in  such  cases. 

7.  Find  two  numbers  whose  sum  is  38  and  whose  difference 
is  50.     Solve  also  without  the  formula. 

8.  The  sum  of  two  numbers  is  48,  and  one  is  3  times  the 
other.     Find  the  numbers. 

9.  The  sum  of  two  numbers  is  108,  and  one  is  6  times  the 
other.     Find  the  numbers. 


PROBLEMS   ON  NUMBER   RELATIONS  113 

10.   The  sum  of  two  numbers  is  s,  and  one  is  k  times  the 
other.     Find  the  numbers. 


Let 

n  =  one  of  the  numbers, 

Then 

s  —  n  —  the  other  number, 

and 

kn  =  s  —  n. 

By  ,4, 

kn  4-  n  =  s. 

By  I, 

(1  +  1)  n  =  s. 

By  A 

n  — ,  one  of  the  nu 

L     l     1 

k  +  1 


and  kn  =k ,  the  other  number. 

*  +  l 

Problems  8  and  9  may  be  solved  by  substitution  in  this 
formula.     State  this  formula  in  words. 

11.  The  sum  of  two  numbers  is  195,  and  one  is  14  times  the 
other.     Find  the  numbers.     Solve  also  without  the  formula. 

12.  The  sum  of  two  numbers  is  75,  and  one  is  f  of  the  other. 
Find  the  numbers.      Solve  also  without  the  formula. 

13.  The  sum  of  two  numbers  is  —  52,  and  one  is  12  times  the 
other.     Find  the  numbers.     Solve  also  without  the  formula. 

14.  If  9  be  added  to  a  number  and  the  sum  multiplied  by  4, 
the  product  equals  7  times  the  number.    What  is  the  number  ? 

15.  If  21  be  added  to  a  number  and  the  sum  multiplied  by  5, 
the  product  equals  12  times  the  number.    What  is  the  number  ? 

16.  If  a  be  added  to  a  number  and  the  sum  be  multiplied  by 
b,  the  product  is  c  times  the  number.     What  is  the  number  ? 

If  n  =  the  number.     Show  that 

ab 
c-b' 

17.  If  24  be  added  to  a  number  and  the  sum  multiplied  by  3, 
the  product  is  9  times  the  number.  Find  the  number.  Solve 
by  use  of  the  formula  of  16,  and  also  without  it. 


114  THE  SOLUTION  OF  PROBLEMS 

18.  If  3  be  added  to  a  number  and  the  sum  multiplied  by 
16,  the  product  is  10  times  the  number.  Find  the  number  by- 
use  of  the  formula,  and  also  without  it. 

19.  The  sum  of  three  numbers  is  108.  The  second  is  16 
greater  than  the  first,  and  the  third  25  greater  than  the  sec- 
ond.    What  are  the  numbers  ? 

20.  The  sum  of  three  numbers  is  98.  The  second  is  7 
greater  than  the  first,  and  the  third  is  9  greater  than  the  sec- 
ond.    What  are  the  numbers  ? 

21.  The  sum  of  three  numbers  is  s.  The  second  is  a  greater 
than  the  first,  and  the  third  is  b  greater  than  the  second. 
What  are  the  numbers? 

. n      l 

If  n  =  the  first  number,  show  that  n=  - — — - — -• 

3 

Translate  the  result  of  problem  21  into  words.  It  should  be  real- 
ized that  if  in  problems  19  and  20  the  numbers  had  been  kept  iincora- 
bined,  as  they  were  necessarily  in  21,  those  results  would  translate 
into  words,  exactly  as  in  21. 

Use  the  formula  derived  in  21  to  solve  the  following,  and 
also  solve  each  one  without  the  formula. 

22.  The  sum  of  three  numbers  is  198.  The  second  is  28 
larger  than  the  first,  and  the  third  is  25  larger  than  the  second. 
What  are  the  numbers  ? 

23.  The  sum  of  three  numbers  is  31.  The  second  is  3  more 
than  the  first,  and  the  third  2  less  than  the  second.  What  are 
the  numbers  ? 

Since  the  third  number  is  2  less  than  the  second,  this  means  that 
b  of  the  formula  is  negative ;  i.e.  b  =  —  2. 

24.  The  sum  of  three  numbers  is  91.  The  second  is  11  less 
than  the  first,  and  the  third  is  12  more  than  the  second.  What 
are  the  numbers  ? 

25.  The  sum  of  three  numbers  is  69.  The  second  is  9  less 
than  the  first,  and  the  third  is  6  less  than  the  second.  What 
are  the  numbers  ?    (How  does  the  formula  apply  in  this  case  ?) 


PROBLEMS  INVOLVING  MOTION  115 

26.  Divide  the  number  248  into  two  parts,  such  that  7  times 
the  first  is  42  more  than  twice  the  second. 

27.  Divide  the  number  645  into  two  parts,  such  that  13  times 
the  first  part  is  20  more  than  6  times  the  other. 

28.  Divide  the  number  a  into  two  parts,  such  that  b  times 
the  first  part  is  c  more  than  d  times  the  second  part. 

If  x  is  the  first  part,  show  that  x  =  fl  • 

b+d 

29.  Translate  the  formula  in  28  into  words. 

30.  Divide  the  number  1240  into  two  parts  such  that  5  times 
the  first  is  200  more  than  the  second.  Solve  by  use  of  the 
formula,  and  also  without  it. 

PROBLEMS   INVOLVING   MOTION 

96.  Problems  like  those  in  the  preceding  class  are  useful 
chiefly  in  cultivating  skill  in  deducing  formulas,  and  so  mak- 
ing rules.  Those,  however,  in  this  and  most  of  the  following 
classes  are  extremely  important  in  themselves,  because  of  the 
simple  laws  of  nature  which  they  exemplify,  and  of  which  they 
afford  a  wide  range  of  application. 

97.  In  scientific  language  the  distance  passed  over  by  a 
moving  body  is  called  the  space,  and  the  number  of  units  of 
space  traversed  is  represented  by  s.  The  rate  of  motion,  that 
is  the  number  of  units  of  space  traversed  in  the  unit  of  time,  is 
called  the  velocity,  and  is  represented  by  v.  The  number  of 
units  of  time  occupied  is  represented  by  t. 

E.g.  At  a  certain  temperature  sound  travels  1080  feet  per  second. 
Hence,  in  5  seconds  it  will  travel  5  •  1080  =  5400  feet.  In  this  case, 
s  =  5400  feet,  v  =  1080  feet  per  second,  t  =  5  seconds. 

We  then  have  the  formula 

*  =  Kf  (1) 


116  THE  SOLUTION  OF  PROBLEMS 

Solve  the  equation  s  =  vt  for  t  in  terms  of  s  and  v,  and  for  v 
in  terms  of  s  and  t. 

Translate  each  of  these  formulas  into  words. 

In  equation  (1)  give  particular  values  to  any  two  of  the  letters  and 
find  the  value  of  the  remaining  one. 

It  is  to  be  understood  in  all  problems  here  considered  that  the  veloc- 
ity remains  the  same  throughout  the  period  of  motion ;  e.g.  sound 
travels  just  as  far  in  any  one  second  as  in  any  other  second  of  its 
passage. 

1.  If  sound  travels  1080  feet  per  second,  how  far  does  it 
travel  in  6  seconds  ? 

2.  If  a  glacier  moves  450  feet  per  year,  how  far  does  it 
move  in  7  years? 

3.  If  a  transcontinental  train  averages  35  miles  per  hour, 
how  far  does  it  travel  in  2£  days  ?  (Given  v  =  35,  t  =  2\  •  24, 
to  find  s.) 

4.  A  hound  runs  23  yards  per  second  and  a  hare  21  yards 
per  second.  If  the  hound  starts  79  yards  behind  the  hare, 
how  long  will  it  require  to  overtake  the  hare  ? 

If  t  is  the  number  of  seconds  required,  then  by  formula  (1)  during 
this  time  the  hound  runs  23 1  yards  and  the  hare  runs  21 1  yards.  Since 
the  hound  must  run  79  yards  farther  than  the  hare,  we  have  : 
23 t  =  21 1  +  79. 

5.  An  ocean  liner  making  21  knots  an  hour  leaves  port  when 
a  freight  boat  making  8  knots  an  hour  is  already  1240  knots 
out.     In  how  long  a  time  will  the  liner  overtake  the  freight  ? 

6.  A  motor  boat  starts  7|  miles  behind  a  sailboat  and  runs 
11  miles  per  hour  while  the  sailboat  makes  6^  miles  per  hour. 
How  long  will  it  require  the  motor  boat  to  overtake  the  sail- 
boat? 

7.  A  freight  train  running  25  miles  an  hour  is  200  miles 
ahead  of  an  express  train  running  45  miles  an  hour.  How  long 
before  the  express  will  overtake  the  freight  ? 


PROBLEMS  INVOLVING  MOTION  117 

8.  A  bicyclist  averaging  12  miles  an  hour  is  52  miles  ahead 
of  an  automobile  running  20  miles  an  hour.  How  soon  will  the 
automobile  overtake  him  ? 

9.  A  and  B  run  a  mile  race.  A  runs  18  feet  per  second 
and  B  Yl\  feet  per  second.  B  has  a  start  of  30  yards.  In 
how  many  seconds  will  A  overtake  B  ?  Which  will  win  the 
race  ? 

10.  Two  objects,  A  and  B,  move  in  the  same  direction,  A  at 
vx*  feet  per  second  and  B  at  v2  feet  per  second.  If  A  has  n  feet 
the  start,  in  how  many  seconds  will  B  overtake  him  ? 

If  t  is  the  number  of  seconds,  then  during  this  time  A  moves  vxt 
feet  and  B  moves  v<£  feet.     Since  B  must  move  n  feet  farther  than  A, 

wehave  v2t=v1t  +  n.  (2) 

The  solution  of  (2)  for  t  gives  the  time  sought. 

It  is  of  the  utmost  importance  that  formulas  (1)  and  (2)  be 
clearly  understood  since  they  are  fundamental  in  every  motion 
problem  in  this  chapter. 

11.  A  fleet,  making  11  knots  per  hour,  is  1240  knots  from 
port  when  a  cruiser,  making  19  knots  per  hour,  starts  out  to 
overtake  it.     How  long  will   it   require  ? 

12.  In  how  many  minutes  does  the  minute  hand  of  a  clock 
gain  15  minute  spaces  on  the  hour  hand  ? 

Using  one  minute  space  for  the  unit  of  distance  and  1  minute  as 
the  unit  of  time,  the  rates  are  1  and  TJ2  respectively,  since  the  hour  hand 
goes  fa  of  a  minute  space  in  1  minute.     Letting  t  be  the  number  of 
minutes  required,  we  have,  just  as  in  problem  10, 
1  •  t  =  fa  t  +  15. 

13.  In  how  many  minutes  after  4  o'clock 
will  the  hour  and  minute  hands  be  together  ? 
(Here  the  minute  hand  must  gain  20  minute 
spaces.) 

*  Vi  and  v2,  read  v  one  and  v  two,  are  used  instead  of  two  different  letters  to 
represent  the  velocities  of  the  first  and  second  respectively. 


118  THE  SOLUTION  OF  PROBLEMS 

14.  At  what  time  between  5  and  6  o'clock  is  the  minute 
hand  15  minute  spaces  behind  the  hour  hand  ?  At  what  time 
is  it  15  minute  spaces  ahead  ? 

Since,  at  4  o'clock,  it  is  25  minute  spaces  behind  the  hour  hand,  in 
the  first  case  it  must  gain  25  —  15  =  10  minute  spaces,  and  in  the 
second  case  it  must  gain  25  +  15  =  40  minute  spaces.  Make  a  dia- 
gram as  in  the  preceding  problem  to  show  both  cases. 

15.  At  what  time  between  9  and  10  o'clock  is  the  minute 
hand  of  a  clock  30  minute  spaces  behind  the  hour  hand  ?  At 
what  time  are  they  together  ? 

In  each  case,  starting  at  9  o'clock,  how  much  has  the  minute  hand 
to  gain  ? 

16.  A  fast  freight  leaves  Chicago  for  New  York  at  8.30  a.m. 
averaging  32  miles  per  hour.  At  2.30  p.m.  a  limited  express 
leaves  Chicago  over  the  same  road,  averaging  55  miles  per 
hour.  In  how  many  hours  will  the  express  overtake  the 
freight? 

If  the  express  requires  t  hours  to  overtake  the  freight,  the  latter 
had  been  on  the  way  t  +  6  hours.  Then  the  distance  covered  by  the 
express  is  55  /,  and  the  distance  covered  by  the  freight  is  32  (t  +  6). 
As  these  must  be  equal,  we  have  55 1  =  32  (t  +  6). 

17.  In  a  century  bicycle  race  one  rider  averages  19^  miles 
per  hour,  while  another,  starting  40  minutes  later,  averages 
22^  miles  per  hour.  In  how  long  a  time  will  the  latter  over- 
take the  former  ? 

18.  A  sparrow  flies  135  feet  per  second  and  a  hawk 
149  feet  per  second.  The  hawk  in  pursuing  the  sparrow 
passes  a  certain  point  7  seconds  after  the  sparrow.  In  how 
many  seconds  from  this  time  does  the  hawk  overtake  the 
sparrow  ? 

19.  A  courier  starts  from  a  certain  point  traveling  vx  miles 
per  hour,  and  a  hours  later  a  second  courier  starts,  going  at 


PROBLEMS  INVOLVING  MOTION  119 

the  rate  of  v2  miles  per  hour.     In  how  long  a  time  will  the 
second  overtake  the  first,  supposing  v2  greater  than  Vi  ? 

If  the  second  courier  requires  t  hours  to  overtake  the  first  the  latter 
had  been  on  the  way  t  +  a  hours.  Thus  the  distance  covered  by  the 
second  courier  is  v%t  and  by  the  first  Vi(t  +  a).  As  these  numbers  are 
equal  we  have  rrf«r,(f-M)  (3) 

20.  In  an  automobile  race  A  drives  his  machine  at  an 
average  rate  of  53  miles  per  hour,  while  B,  who  starts  \  hour 
later,  averages  57  miles  per  hour.  How  long  does  it  require  B 
to  overtake  A  ?  Use  formula  (3).  Solve  also  by  finding  how 
far  A  has  gone  when  B  starts  and  then  use  formula  (2). 

21.  A  freight  steamer  leaves  New  York  for  Liverpool  aver- 
aging 10^  knots  per  hour,  and  is  followed  4  days  later  by  an 
ocean  greyhound  averaging  20^  knots  per  hour.  In  how  long 
a  time  will  the  latter  overtake  the  former  ? 

22.  One  athlete  makes  a  lap  on  an  oval  track  in  26  seconds, 
another  in  28  seconds.  If  they  start  together  in  the  same 
direction,  in  how  many  seconds  will  the  first  gain  one  lap  on 
the  other  ?     Two  laps  ? 

Let  one  lap  be  the  unit  of  distance.  Since  the  first  covers  one  lap 
in  26  seconds  his  rate  per  second  is  fa.  Likewise  the  rate  of  the  other 
is  fa.  If  t  is  the  required  number  of  seconds  the  distance  covered  by 
the  first  is  ^g  t  and  by  the  second  fa t.  If  the  first  goes  one  lap  farther 
than  the  second  the  equation  is  -fat  =  fat  +  1;  if  two  laps  farther  it 
is  fa  t  =  fa  t  +  2. 

23.  Two  automobiles  are  racing  on  a  circular  track.  One 
makes  the  circuit  in  31  minutes  and  the  other  in  38£  minutes. 
In  what  time  will  the  faster  machine 

gain  1  lap  on  the  slower  ?  y^ ,>v. 

24.  The  planet  Mercury  makes  a  cir-  /  jr  >v  \ 
cuit  around  the  sun  in  3  months  and  /    /     sun^.^A.lvenua 
Venus  in  1\  months.     Starting  in  con-  \    V  /    / 
junction,  as  in  the  figure,  how  long  be-  \    v^,    ^S / 
fore  they  will  again  be  in  this  position  ?         ^s.^     ^y^ 


120  THE  SOLUTION   OF  PROBLEMS 

25.  Saturn  goes  around  the  sun  in  29  years  and  Jupiter  in 
12  years.  Starting  in  conjunction,  how  soon  will  they  be  in 
conjunction  again  ? 

26.  Uranus  makes  the  circuit  of  its  orbit  in  84  years  and 
Neptune  in  164  years.  If  they  start  in  conjunction,  how  long 
before  they  will  be  in  conjunction  again? 

27.  The  hour  hand  of  a  watch  makes  one  revolution  in  12 
hours,  and  the  minute  hand  in  one  hour.  How  long  is  it  from  the 
time  when  the  hands  are  together  until  the}'  are  again  together  ? 

28.  One  object  makes  a  complete  circuit  in  a  units  of  time 
and  another  in  b  units  (of  the  same  kind).  In  how  many  units 
of  time  will  one  overtake  the  other,  supposing  b  to  be  greater 
than  a  ? 

29.  At  what  times  between  12  o'clock  and  6  o'clock  are  the 
hands  of  a  watch  together  ?  (Find  the  time  required  to  gain 
one  circuit,  two  circuits,  etc.) 

PROBLEMS  INVOLVING  THE  LEVER 

98.    Two  boys,  A  and  B,  play  at  teeter.     They  find  that  the 

teeter  board  will  balance  when  equal  products  are  obtained  by 

multiplying  the  weight  of  each  by  his  distance  from  the  point 

of  support. 

Thus,  if  B  weigh? 

U-!0-^     _     _  A____     __     ywih)       80   pounds    and   is    t 

•  « £5   "  ~IJ5    '  feet    from    the    poinfc 

of  support,  then  A,  who  weighs  100  lbs.,  must  be  4  feet  from  this 
point,  since   80  x  5  =  100  x  4. 

The  teeter  board  is  a  certain  kind  of  lever;  the  point  of 
support  is  called  the  fulcrum. 

In  each  of  the  following  problems  make  a  diagram  similar 
to  the  above  figure  : 

1.   A  and  B  weigh  90  and  105  lbs.  respectively.      If  A  is 
seated  7  feet  from  the  fulcrum,  how  far  is  B  from  this  point  ? 


PROBLEMS  INVOLVING   THE  LEVER  121 

2.  Using  the  same  weights  as  in  the  preceding  problem,  if 
B  is  6^  feet  from  the  fulcrum,  how  far  is  A  from  that  point  ? 

3.  A  and  B  are  5  and  7  feet  respectively  from  the  fulcrum. 
If  B  weighs  75  pounds,  how  much  does  A  weigh  ? 

4.  A  and  B  weigh  100  and  110  pounds  respectively.  A 
places  a  stone  on  the  board  with  him  so  that  they  balance 
when  B  is  6  feet  from  the  fulcrum  and  A  5^  feet  from  this 
point.     How  heavy  is  the  stone  ? 

5.  If  the  distances  from  the  boys  to  the  fulcrum  are  respec- 
tively dv  and  d2,  and  their  weights  w^  and  w2,  then 

dlwl  =  d2w2.  (1) 

If  any  three  of  these  four  numbers  are  given,  the  fourth  may  be 
found  by  means  of  this  equation.  Solve  dxivx  =  d2w2  for  each  of  the 
four  numbers  involved  in  terms  of  the  other  three.  Problems  1  to  4 
can  be  solved  by  substitution  in  the  formulas  thus  obtained. 

6.  A  and  B  are  seated  at  the  opposite  ends  of  a  13-foot  teeter 
board.  Using  the.  weights  of  problem  1,  where  must  the  ful- 
crum be  located  so  that  they  shall  balance  ? 

If  the  fulcrum  is  the  distance  d  from  A  then  it  is  (13  —  d)  from  B. 
Hence  90tf  =  105  (13  -d). 

7.  A  and  B  together  weigh  212^  pounds.  They  balance 
when  A  is  6  feet,  and  B  6f  feet,  from  the  fulcrum.  Find  the 
weight  of  each. 

8.  A,  who  weighs  75  pounds,  sits  7  feet  from  the  fulcrum,  and 
B,  who  weighs  105  pounds,  sits  on  the  other  side.  At  what  dis- 
tance from  the  fulcrum  should  B  sit  in  order  to  make  a  balance  ? 

9.  If  in  the  preceding  problem  C,  weighing  48  pounds,  sits  on 
the  side  with  A  and  4  feet  from  the  fulcrum,  where  must  D, 
who  weighs  64  pounds,  sit  so  as  to  maintain  a  balance? 

From  problem  8,  when  the  teeter  balanced  it  was  found  that  A  'a 
weight  acted  like  a  lever  with  a  downward  force  of  7  •  75  pounds  and 
B's  on  the  other  side  with  a  force  of  5  •  105  pounds. 

7  •  75  =  5  •  105. 


122  THE  SOLUTION  OF  PROBLEMS 

Then  in  problem  9,  C  and  D  must  sit  so  as  to  keep  the  board  in 
balance,  that  is,  so  as  to  add  the  same  downward  force  to  both  sides. 
Hence,  as  4-48  is  added  on  the  side  of  A,  3  •  64  must  be  added  on 
the  side  of  B,  since  4  •  48  =  3  •  04.  Therefore,  adding  these  two 
equations,  we  still  have  the  balance,  namely, 

7  •  75  +  4  •  48  =  5  •  105  +  3  •  64. 

B  (105  lbs.)        D(Ulbs.) (7(18 lbs)  (75  lbs.)  A. 

L | 3-feer A J-Tevr '  ] 

6  feet.  ^m  7  feet 

The  weight  of  the  boy  multiplied  by  his  distance  from  the 
fulcrum  is  called  his  leverage.  The  sum  of  the  leverages  on 
the  two  sides  must  be  the  same.  Hence,  if  two  boys,  weighing 
respectively  wx  and  w2  pounds,  are  sitting  at  distances  dx  and 
d2  on  one  side,  and  two  boys,  weighing  w3  and  wt  pounds,  sitting 
at  distances  d3,  dt  on  the  other  side,  then 

wxdx  +  wJ,  =  Ms  +  »A-  (2) 

10.  If  two  boys  weighing  75  and  90  pounds  sit  at  distances 
of  3  and  5  feet  respectively  on  one  side  and  one  weighing  82 
pounds  sits  at  3  feet  on  the  other  side,  where  should  a  boy 
weighing  100  pounds  sit  in  order  to  make  the  board  balance  ? 

11.  A  beam  carries  a  weight  of  240  pounds  1\  feet  from  the 
fulcrum  and  a  weight  of  265  pounds  at  the  opposite  end  which 
is  10  feet  from  the  fulcrum.  On  which  side  and  how  far  from 
the  fulcrum  should  a  weight  of  170  pounds  be  placed  so  as  to 
make  the  beam  balance  ? 


PROBLEMS  INVOLVING  DENSITIES 
99.  If  a  cubic  foot  of  a  certain  kind  of  rock  weighs  2.5  times 
as  much  as  a  cubic  foot  of  fresh  water,  the  density  of  this  rock 
is  said  to  be  2.5.  If  a  cubic  foot  of  oak  weighs  .85  times  as 
much  as  a  cubic  foot  of  fresh  water,  the  density  of  the  oak  is 
.85.  The  density  of  water  is  taken  as  the  standard  with  which 
the  densities  of  other  substances  are  compared. 


PROBLEMS  INVOLVING  DENSITIES  123 

Thus,  when  we  say  that  the  density  of  a  certain  kind  of  iron  is 
7.25,  we  mean  that  any  given  volume  of  the  iron  weighs  7.25  times  as 
much  as  a  like  volume  of  fresh  water  ;  and  when  we  say  that  the 
density  of  cork  is  .24,  we  mean  that  a  given  volume  of  cork  weighs 
.24  times  as  much  as  a  like  volume  of  fresh  water. 

A  cubic  centimeter  of  distilled  water  at  the  freezing  point 
which  weighs  one  gram  is  used  as  a  standard  of  comparison. 
We  therefore  say  that  the  density  of  any  substance  is  equal  to 
the  number  of  grams  which  a  cubic  centimeter  of  it  weighs. 

E.g.  if  a  cubic  centimeter  (ccm.)  of  a  certain  kind  of  marble 
weighs  2.5  grams,  then  it  weighs  2.5  times  as  much  as  the  same  vol- 
ume of  water,  and  hence  its  density  is  2.5. 

The  weight  of  an  object  in  grams  is,  therefore,  the  product 
of  its  volume  in  cubic  centimeters  multiplied  by  its  density. 

E.g.  if  the  density  of  cork  is  .24,  this  means  that  a  cubic  centi- 
meter of  cork  weighs  .24  grams.  Any  volume  of  cork,  say  10  ccm., 
weighs  10  x  .24  =  2.4  grams.  Hence,  if  we  represent  the  weight  of  an 
object  in  grams  by  w,  its  volume  in  cubic  centimeters  by  v,  and  its 
density  by  d,  we  have  the  relation, 

w  =  vd.  (1) 

1.  What  is  the  weight  of  an  object  whose  density  is  4.3  and 
whose  volume  is  250  ccm.  ?     (Here  v  =  250,  d  =4.3.    Find  w.) 

2.  What  is  the  density  of  an  object  whose  weight  is  23.5 
and  whose  volume  is  17  ccm.  ? 

3.  What  is  the  volume  of  an  object  whose  weight  is  24 
grams  and  whose  density  is  .65  ? 

4.  If  500  ccm.  of  alcohol,  density  .79,  is  mixed  with  300 
ccm.  of  distilled  water,  what  is  the  density  of  the  mixture  ? 

The  volume  of  the  mixture  is  the  sum  of  the  volumes,  and  the 
weight  of  the  mixture  is  the  sum  of  the  weights,  of  the  water  and 
alcohol.     Hence,  from  formula  (1)  : 

500  x  .79  +  300  x  1  =  d  x  800.     To  find  d. 


124  THE  SOLUTION  OF  PROBLEMS 

5.  If  1200  ccm.  of  cork,  density  .24,  are  combined  with  64 
ccm.  of  steel,  density  7.8,  what  is  the  average  density  of  the 
combined  mass  ?  Will  it  float  or  sink  ?  (A  substance  sinks 
if  its  density  is  greater  than  that  of  water.) 

In  solving  problems  of  this  kind  find  the  expressions  representing 
weight,  density,  and  volume,  and  substitute  in  the  equation  w  =  dv. 

6.  How  many  cubic  centimeters  of  cork,  density  .24,  must 
be  combined  with  75  ccm.  of  steel,  density  7.8,  in  order  that 
the  average  density  shall  be  equal  to  that  of  water,  i.e.  so  that 
the  combined  mass  will  just  float  ? 

Let  v  =  volume  of  cork  to  be  used.  Then  the  total  volume  is 
75  +  v,  the  total  weight  is  75  x  7.8  +  .24  v,  and  the  density  is  1. 
Hence,  75  x  7.8  +  .24  v  =  1  •  (75  +  r).     Solve  this  equation  for  v. 

7.  Brass  is  an  alloy  of  copper  and  zinc.  How  many  cubic 
centimeters  of  zinc,  density  6.86,  must  be  combined  with  100 
ccm.  of  copper,  density  8.83,  to  form  brass  whose  density  is 
8.31? 

8.  Coinage  silver  is  an  alloy  of  copper  and  silver.  How 
many  ccm.  of  copper,  density  8.83,  must  be  added  to  10  ccm.  of 
silver,  density  10.57,  to  form  coinage  silver,  whose  density  is 
10.38  ? 

9.  The  density  of  pure  gold  is  19.36  and  of  nickel  8.57. 
How  many  ccm.  of  nickel  must  be  mixed  with  10  ccm.  of  pure 
gold  to  form  14  karat  gold  whose  density  is  14.88. 

10.  How  much  mercury,  density  13.6,  must  be  added  to  20 
ccm.  of  gold,  density  19.36,  so  that  the  density  of  the  compound 
shall  be  16.9  ? 

11.  What  is  the  average  density  of  40  ccm.  of  water,  den- 
sity 1,  and  180  ccm.  of  alcohol,  density  .79  ? 

12.  How  many  cubic  centimeters  of  water  must  be  mixed 
with  350  ccm.  of  alcohol,  so  that  the  density  of  the  mixture 
shall  be  .97? 


PROBLEMS   ON  MOMENTUM  125 

13.  The  density  of  copper  is  8.83.  500  ccm.  of  copper  is 
mixed  with  700  ccm.  of  lead,  whose  density  is  11.35.  What  is 
the  density  of  the  combined  mass  ? 

14.  "When  960  ccm.  of  iron,  density  7.3,  is  fastened  to  8400 
ccm.  of  white  pine,  the  combination  just  floats,  i.e.  has  a  density 
of  1.     What  is  the  density  of  white  pine  ? 


PROBLEMS   ON   MOMENTUM 

100.  The  force  with  which  a  moving  body  strikes  another 
depends  both  upon  its  weight  and  upon  its  rate  of  motion. 
The  product  of  the  weight  and  velocity  of  a  moving  body  is 
called  its  momentum.  The  weight  of  a  body  is  also  commonly 
called  its  mass. 

What  is  the  momentum  of  a  body  whose  weight  is  10  pounds  and 
which  moves  15  feet  per  second  ? 

What  is  the  velocity  of  a  body  whose  weight  is  50  and  whose 
momentum  is  350  pounds  ? 

What  is  the  weight  of  a  body  whose  momentum  is  500  and  whose 
velocity  is  25  feet  per  second  ? 

A  bullet  weighing  \  of  a  pound,  moving  2250  feet  per  second, 
has  a  momentum  equal  to  that  of  a  stone  weighing  50  pounds,  which 
is  hurled  at  the  rate  of  9  feet  per  second,  since  \  •  2250  =  50  •  9. 

By  careful  experiment  it  has  been  found  that  when  a  mov- 
ing body  strikes  a  body  at  rest  but  free  to  move,  the  two  will 
move  on  with  a  combined  momentum  equal  to  the  momen- 
tum of  the  first  body  before  the  impact. 

Thus,  if  a  freight  car,  weighing  25  tons  and  moving  at  the  rate  of 
12  miles  per  hour,  strikes  a  standing  car  weighing  15  tons,  the  two  will 
move  on  with  the  original  momentum  of  12  •  25.  But  as  the  combined 
weight  is  now  25  +  15,  the  rate  of  motion  has  been  decreased  to  7\ 
miles  per  hour,  since  12  •  25  =  1\  (25  +  15).  In  this  case  the  automatic 
coupler  connects  the  cars  and  they  move  on  together.    Even  if  the 


126  THE  SOLUTION   OF  PROBLEMS 

two  bodies  after  impact  do  not  cling  together,  as  two  croquet  balls, 
still  the  momentum  of  the  one  plus  that  of  the  other  equals  the 
original  momentum. 

E.g.  if  a  croquet  ball  weighing  8  ounces  and  moving  20  feet  per 
second,  strikes  another  weighing  7  ounces  and  starts  it  off  at  the 
rate  of  18  feet  per  second,  then  if  the  diminished  velocity  of  the 
first  ball  is  called  v,  we  have 

8-20  =  7-  18  +  8  v, 
and  solving,  v  —  4.25. 

This  indicates  that  the  first  ball  is  nearly  stopped,  which  coincides 
with  common  observation. 

1.  In  a  switch  yard  a  car  weighing  40  tons  and  moving  8 
miles  per  hour  strikes  a  standing  car  weighing  24  tons.  What 
is  the  velocity  of  the  two  after  impact  ? 

2.  A  billiard  ball  weighing  6  ounces  and  moving  16  feet 
per  second  strikes  another  ball  which  it  sends  off  at  the  rate  of 
10  feet  per  second.  The  rate  of  the  first  ball  is  reduced  to  9 
feet  per  second  by  the  impact.  What  is  the  weight  of  the 
second  ball  ? 

Since  the  momentum  before  impact  equals  the  sum  of  the  mo- 
mentums  after  impact,  we  have  6  •  16  =  9  •  6  +  10  w,  w  being  the 
unknown  weight  of  the  second  ball. 

3.  A  bowler  uses  a  16-ounce  ball  to  take  down  the  last  pin. 
The  ball  sends  the  pin  off  at  a  velocity  of  6  feet  per  second, 
the  weight  of  the  pin  being  48  ounces,  while  the  velocity  of 
the  ball  is  reduced  to  4  feet  per  second.  With  what  velocity 
did  the  ball  strike  the  pin  ? 

Since  the  momentum  before  impact  equals  the  sum  of  the  mo- 
mentums  after  impact,  we  have  16 1>  =  6  -48  +  4- 16,  v  being  the 
unknown  velocity  of  the  ball  before  impact. 

4.  In  each  of  these  problems  we  have  considered  the  weight 
of  two  bodies,  which  we  may  call  wx  and  w2.  If  we  call  vx  the 
velocity  with  which  the  first  strikes  the  second,  v2  the  velocity 


PROBLEMS   ON  MOMENTUM  127 

imparted  to  the  second,  and  v/  the  resulting  decreased  velocity 
of  the  first,*  we  have 

■W  =  Wjft'  +  w2  v2  (1) 

Translate  this  equation  into  words.  It  contains  5  different  num- 
bers, toi,  w2,  vi,  vi',  and  v2.  If  any  four  of  these  are  given,  the  fifth 
may  be  found  by  solving  this  equation  for  that  one  in  terms  of  the 
other  four. 

5.  Solve  the  equation  (1)  for  vx  in  terms  of  w1}  w^  v/,  and 
v2 .     Translate  into  words. 

6.  Solve  the  equation  (1)  above  for  Wy  in  terms  of  vi}  vx',  w2> 
and  v2.     Translate  into  words. 

7.  Solve  the  equation  (1)  above  for  vx'  in  terms  of  wv  iv2, 
vv  and  Vjj.     Translate  into  words. 

8.  Solve  equation  (1)  for  w2  in  terms  of  wv  v1}  V>  v2?  an(i 
translate  the  result  into  words. 

9.  Solve  equation  (1)  for  v2  in  terms  of  w1}  w2,  vlt  v^,  and 
translate  the  result  into  words. 

10.  In  the  result  of  the  last  exercise  substitute  w1  =  50, 
w2  =  40,  vx  =  10,  Vy  =  2,  and  find  the  value  of  v2.  Make  a 
problem  to  fit  this  case. 

11.  In  the  result  of  problem  8  substitute  wx  =  1000, 
V!=75,  Vi  =  25,  v2  =  50,  and  find  the  value  of  w2.  Make  a 
problem  to  fit  this  case. 

12.  In  the  result  of  problem  6  substitute  vx  =  60,  V  =  10, 
w2  =  80,  v2  =  25,  and  find  the  value  of  wx.  Make  a  problem 
to  fit  this  case. 

13.  In  the  result  of  problem  7  substitute  v\  =  250,  w2  =  125, 
Vy  =  50,  and  v2  =  50,  and  find  the  value  of  vx\  Make  a  prob- 
lem to  fit  this  case. 

*  v-l  is  read  v  one  prime  and  is  used  rather  than  a  new  letter  for  the 
purpose  of  recalling  that  this  is  another  rate  of  the  first  body. 


128 


THE  SOLUTION  OF  PROBLEMS 


PROBLEMS  ON  THERMOMETER  READINGS 

101.  There  are  two  kinds  of  thermometers  in  use  in  this 
country,  called  the  Fahrenheit  and  Centigrade,  the  former  for 
common  purposes,  and  the  latter  for  scientific 
records  and  investigations.  Hence  it  frequently 
becomes  necessary  to  translate  readings  from 
one  kind  to  the  other. 

The  freezing  and  boiling  points  are  two 
fixed  temperatures  by  means  of  which  the  com- 
putations are  made.  On  the  Centigrade  these 
are  marked  0°  and  100°  respectively  and  on  the 
Fahrenheit  they  are  marked  32°  and  212°  respec- 
tively. See  the  cut.  Hence  between  the  two 
fixed  points  there  are  100  degrees  Centigrade 
and  180  degrees  Fahrenheit. 

That  is,  100  degree  spaces  on  the  Centigrade  cor- 
respond to  180  degree  spaces  on  the  Fahrenheit. 

Hence  1°  Centigrade  corresponds  to  |°  Fahrenheit, 
or  1°  Fahrenheit  corresponds  to  |°  Centigrade. 

All  problems  comparing  the  two  thermometers  are  solved  by 
reference  to  these  fundamental  relations. 

1.  If  the  temperature   falls   15   degrees   Centigrade,   how 
many  degrees  Fahrenheit  does  it  fall  ? 

2.  If  the   temperature  rises   18   degrees   Fahrenheit,  how 
many  degrees  Centigrade  does  it  rise  ? 

3.  Translate  +  25°  Centigrade  into  Fahrenheit  reading. 

25°  Centigrade  equals  |  •  25°  =  45°  Fahrenheit. 
45°  above  the  freezing  point  =  45°  +  32°    above  0°  Fahrenheit. 
Hence,  calling  the  Fahrenheit  reading  F,  we  have  F  =  32  +  f  •  25. 

4.  Translate  +  14°  Centigrade  into  Fahrenheit  reading. 
Eeasoning  as  before,  F  =  32  +  §  •  14. 


PROBLEMS   ON   THERMOMETER   READINGS        129 

From  the  two  preceding  problems  we  have  the  formula 

F  =  32  +  |C.  (1) 

Translate  this  into  words,  understanding  that  F  and  C  stand 
for  readings  on  the  respective  thermometers. 

5.  Solve  the  above  equation  for  C  in  terms  of  F  and  find 

C  =  f(F-32).  (2) 

Translate  this  into  words.  Verify  the  solution  of  each  of 
the  following  by  referring  to  the  cut  on  page  128. 

6.  Translate   -f  41°  Fahrenheit   into   Centigrade   reading. 
Substitute  in  the  proper  formula. 

7.  Translate  +  98°  Fahrenheit,  blood  heat,  into  Centigrade 
reading. 

8.  Translate  —  20°  Fahrenheit  into  Centigrade  reading. 

9.  Translate  —  40°  Centigrade,  the  freezing  point  of  mer- 
cury, into  Fahrenheit  reading. 

10.  Translate  0°  Fahrenheit  into  Centigrade  by  use  of  the 
formula.     Also  0°  Centigrade  into  Fahrenheit. 

11.  Translate  -f-  212°  Fahrenheit  into  Centigrade,  and  also 
+ 100°  Centigrade  into  Fahrenheit. 

12.  What  is  the  temperature  Centigrade  when  the  sum  of 
the  Centigrade  and  Fahrenheit  readings  is  102°  ? 

13.  What  is  the  temperature  Fahrenheit  when  the  sum  of 
the  Centigrade  and  Fahrenheit  readings  is  zero  ? 

14.  What  is  the  temperature  Centigrade  when  the  sum  of 
the  Centigrade  and  Fahrenheit  readings  is  140°? 

15.  What  is  the  temperature  in  each  reading  when  the 
Fahrenheit  is  50°  higher  than  the  Centigrade  ? 

16.  The  Fahrenheit  reading  at  the  temperature  of  liquid 
air  is  128  degrees  lower  than  the  Centigrade  reading.  Find 
both  the  Centigrade  and  the  Fahrenheit  reading  at  this 
temperature. 


130  THE  SOLUTION  OF  PROBLEMS 

PROBLEMS  ON   THE   ARRANGEMENT   AND   VALUE   OF   DIGITS 

102.  If  we  speak  of  the  number  whose  3  digits,  in  order 
from  left  to  right,  are  5,  3,  and  8,  we  mean  538  =  500  -f  30  +  8. 
Likewise,  the  number  whose  three  digits  are  h,  t,  and  u  is  writ- 
ten 100  h  +  10  t  +  u. 

Hence,  when  letters  stand  for  the  digits  of  numbers  written 
in  the  decimal  notation,  care  must  be  taken  to  multiply 
each  letter  by  10,  100,  1000,  etc.,  according  to  the  position  it 
occupies. 

Illustrative  Problem.  A  number  is  composed  of  two  digits 
whose  sum  is  6.  If  the  order  of  the  digits  is  reversed,  we 
obtain  a  number  which  is  18  greater  than  the  first  number. 
What  is  the  number  ? 

Solution.     Let  x  =  the  digit  in  tens'  place. 

Then  6  —  x  =  the  digit  in  units'  place. 

Hence,  the  number  is  10  x  +  6  —  x.  Reversing  the  order  of  the 
digits,  we  have  as  the  new  number  10(6  —  x)  +  x. 

Hence  10(6  -  x)  +  x  =  18  +  10  x  +  6  -  x. 

1.  A  number  is  composed  of  two  digits,  the  digit  in  units' 
place  being  2  greater  than  the  digit  in  tens'  place.  If  4  is 
added  to  the  number,  it  is  then  equal  to  5  times  the  sum  of 
the  digits.     What  is  the  number? 

2.  A  number  is  composed  of  two  digits,  the  digit  in  tens' 
place  being  3  greater  than  the  digit  in  units'  place.  The  num- 
ber is  one  more  than  8  times  the  sum  of  the  digits.  What 
is  the  number  ? 

3.  A  number  is  composed  of  two  digits  whose  sum  is  9. 
If  the  order  of  the  digits  is  reversed,  we  obtain  a  number 
which  is  equal  to  12  times  the  remainder  when  the  units'  digit 
is  taken  from  the  tens'  digit.     What  is  the  number  ? 


REVIEW  QUESTIONS  131 

4.  A  number  is  composed  of  two  digits  whose  difference 
is  4.  If  the  order  of  the  digits  is  reversed,  we  obtain  a  num- 
ber which  is  3  less  than  4  times  the  sum  of  the  digits.  What 
is  the  number  ? 

5.  The  digit  in  tens'  place  is  1  more  than  twice  the  digit 
in  units'  place.  If  36  is  subtracted  from  the  number,  the  order 
of  the  digits  will  be  reversed.     What  is  the  number? 

6.  The  digit  in  units'  place  is  2  less  than  twice  the  digit  in 
tens'  place.  If  the  order  of  the  digits  is  reversed,  the  number 
is  unchanged.     What  is  the  number  ? 

7.  The  digit  in  tens'  place  is  12  less  than  5  times  the  digit 
in  units'  place.  If  the  order  of  the  digits  is  reversed,  the  num- 
ber is  equal  to  4  times  the  sum  of  the  digits.  What  is  the 
number  ? 

8.  A  number  is  composed  of  three  digits.  The  digit  in 
units'  place  is  3  greater  than  the  digit  in  tens'  place,  which  in 
turn  is  2  greater  than  the  digit  in  hundreds'  place.  The  num- 
ber is  equal  to  96  plus  4  times  the  sum  of  the  digits.  What  is 
the  number  ? 

REVIEW  QUESTIONS 

1.  In  order  that  a  problem  may  be  solved  by  means  of  a 
formula,  how  many  of  the  letters  in  the  formula  must  be  given 
by  the  problem  ?     Illustrate  this  by  the  formula  i  =prt. 

2.  State  the  formulas  involving  areas  and  volumes  which 
have  been  used  in  this  chapter.  Solve  each  formula  for  each 
of  its  letters  in  terms  of  all  the  others. 

3.  The  area  of  a  circle  is  found  by  squaring  its  radius  and 
multiplying  by  3.1416.  State  this  rule  as  a  formula,  using  the 
Greek  letter  tt  for  3.1416. 

4.  The  volume  of  a  circular  column  is  found  by  multi- 
plying the  area  of  its  base  by  its  height.  State  this  rule  as 
a  formula,  using  r  for  the  radius  of  the  base  and  h  for  the 
height. 


132  THE  SOLUTION  OF  PROBLEMS 

5.  State  formulas  for  finding  two  numbers  when  their  sum 
and  difference  are  given. 

Find  a  number  such  that  if  1G  be  subtracted  from  it,  5  times 
this  difference  equals  the  number. 

Construct  other  problems  like  this  and  then  make  a  formula 
for  the  solution  of  all  such  problems.     (See  page  110.) 

6.  State  three  important  formulas  used  in  the  solution  of 
motion  problems.  Translate  each  into  words.  Solve  each  for- 
mula for  each  of  its  letters. 

Of  the  motion  problems,  4  to  29,  which  ones  involve  formula 
(1)  ?  Which  formula  (2)  and  which  formula  (3)  ? 

7.  State  the  formulas  used  in  this  chapter  in  working 
problems  on  the  lever.  Solve  each  formula  for  each  of 
its  letters. 

Using  a  teeter  board  12  feet  long  with  the  fulcrum  in  the 
middle,  how  may  three  boys  weighing  50,  75,  and  100  lbs.  re- 
spectively be  seated  on  it  so  as  to  make  the  board  balance  ?  Is 
there  more  than  one  solution  ?  Make  a  diagram  for  each  of 
your  results. 

8.  What  is  meant  by  the  density  of  a  substance  ?  What  is 
used  as  a  standard  (unit)  of  density  with  which  the  densities  of 
other  substances  are  compared  ? 

What  is  the  relation  between  the  weight  of  a  substance,  its 
volume  and  its  density  ? 

9.  Define  momentum.  When  a  moving  car  strikes  a  stand- 
ing car,  free  to  move,  what  can  you  say  of  the  momentum  of 
the  two  after  impact? 

10.  Describe  the  Fahrenheit  and  Centigrade  thermometers. 
State  the  formulas  for  the  reading  of  each  thermometer  in  terms 
of  the  other. 

11.  Write  347  as  a  trinomial.  Write  the  number  whose 
three  digits  in  order  are  a,  b,  c ;  also  the  number  whose  three 
digits  in  order  are  c,  b,  a. 


REVIEW  PROBLEMS  133 

REVIEW  PROBLEMS 

103.  Most  of  the  following  problems  can  be  solved  by  sub- 
stitution in  some  of  the  formulas  developed  in  this  chapter. 
Solve  as  many  as  possible  in  this  way. 

1.  One  boy  runs  around  a  circular  track  in  26  seconds,  and 
another  in  30  seconds.  In  how  many  seconds  will  they  again 
be  together,  if  they  start  at  the  same  time  and  place  and  run 
in  the  same  direction  ? 

2.  Divide  the  number  144  into  two  parts,  so  that  ^  of  the 
greater  is  9  more  than  \  of  the  smaller. 

3.  The  sum  of  two  numbers  is  2890.  Seven  times  one 
is  266  less  than  5  times  the  other.     What  are  the  numbers  ? 

4.  What  is  the  simple  interest  on  $  400  at  6|  %  for  7  years 
and  9  months  ? 

5.  The  number  of  telegraph  messages  sent  in  the  United 
States  in  1905  was  5  million  less  than  three  times  as  great  as 
in  1880,  and  69  million  less  than  twice  that  in  1900 ;  while  in 
1900  it  was  16  million  more  than  twice  as  great  as  in  1880. 
Find  the  number  of  messages  in  each  of  these  years. 

6.  The  same  number  is  added  to  each  of  the  numbers  8,  9, 
10, 12.  What  is  this  number  if  the  product  of  the  first  and  last 
sums  is  equal  to  the  product  of  the  second  and  third  sums  ? 

7.  Find  the  time  between  4  and  5  o'clock  when  the  hands 
of  the  clock  are  30  minute  spaces  apart. 

8.  A  man  buys  a  house  for  $6500.  His  yearly  tax  on  the 
property  is  $57.  The  coal  costs  $60  per  year,  repairs  $50, 
and  janitor  service  $108.  To  what  monthly  rental  are  his 
expenses  equivalent  if  money  is  worth  5  %  ? 

9.  A  boatman  rowing  down  a  river  makes  23  miles  in  3 
hours  and  returns  at  the  rate  of  3^  miles  per  hour.  How  swift 
does  the  river  flow  ? 


134  THE  SOLUTION  OF  PROBLEMS 

10.  A  bird  flying  with  the  wind  goes  65  miles  per  hour,  and 
flying  against  a  wind  twice  as  strong  it  goes  20  miles  per  hour. 
What  is  the  rate  of  the  wind  in  each  case  ? 

11.  A  steamer  going  with  the  tide  makes  19  miles  per  hour, 
and  going  against  a  current  \  as  strong  it  makes  13  miles  per 
hour.     What  is  the  speed  of  the  steamer  in  still  water  ? 

12.  A  boatman  trying  to  row  up  a  river  drifts  back  at  the 
rate  of  1^  miles  per  hour,  while  he  can  row  down  the  river  at 
the  rate  of  12  miles  per  hour.    What  is  the  rate  of  the  current? 

13.  A  boatman  rowing  with  the  tide  makes  n  miles  per 
hour ;  rowing  against  a  tide  k  times  as  strong,  he  makes  m  miles 
per  hour.  At  what  rate  does  he  row,  and  what  is  the  velocity 
of  the  stream  ?     State  the  result  as  a  formula. 

14.  A  beam  is  16  feet  long.  •  At  what  point  must  it  be  sup- 
ported if  it  is  to  carry,  when  balanced,  460  pounds  at  one  end 
and  690  at  the  other,  its  weight  being  disregarded  ? 

15.  Two  numbers  differ  by  2  and  the  difference  of  their 
squares  is  100.     What  are  the  numbers  ? 

16.  A  beam  is  12  feet  long.  It  carries  a  40-pound  weight 
at  one  end,  a  60-pound  weight  3  feet  from  this  end,  and  a 
70-pound  weight  at  the  other  end.  Where  is  the  fulcrum  if 
the  beam  is  balanced  ? 

17.  There  is  a  number  composed  of  2  digits  whose  tens'  digit 
is  2  less  than  its  units'  digit.  The  number  is  1  less  than  5  times 
the  sum  of  its  digits.     What  is  the  number  ? 

18.  A  farm  containing  240  acres  can  be  rented  at  $3  per 
acre.  The  renter  finds  that  if  he  borrows  money  at  5  °J0  to 
buy  the  farm  he  will  save  $125  per  year.     Find  its  value. 

19.  Two  trains  start  from  the  same  point  at  the  same  time 
and  in  the  same  direction,  one  making  25  miles  per  hour  and  the 
other  42  miles  per  hour.     When  will  they  be  238  miles  apart  ? 


REVIEW  PROBLEMS  135 

20.  The  number  of  national  banks  in  the  United  States  on 
March  1,  1906,  was  1322  less  than  twice  as  many  as  on 
March  1,  1900,  and  the  number  in  1903  was  1009  greater  than 
in  1900.  If  the  number  in  1906  be  subtracted  from  3  times 
the  number  in  1903  the  remainder  is  7736.  Find  the  number 
in  each  of  the  three  years  mentioned. 

21.  The  earth  and  Mars  were  in  conjunction  July  12,  1907. 
When  are  they  next  in  conjunction  if  the  earth's  period  is 
365  days  and  that  of  Mars  687  days  ?      (See  figure,  page  119.) 

22.  $  7500  is  invested  at  4  %  simple  interest.  Seven  years 
and  three  months  later  the  amount  is  used  to  build  a  house. 
What  is  the  cost  of  the  house  if  $  735  has  to  be  added  to  com- 
plete it  ? 

23.  There  is  a  number  composed  of  two  digits  whose  sum 
is  12.  If  the  order  of  the  digits  is  reversed,  the  number  is 
increased  by  18.     Find  the  number. 

24.  A  slow  steamer  sails  from  New  York  to  Liverpool,  mak- 
ing 9  knots  per  hour.  A  swift  liner  follows  62  hours  later, 
making  20|  knots  per  hour.  In  how  many  hours  will  the  latter 
overtake  the  former? 

25.  A  man  invested  a  certain  sum  of  money  at  5  %  simple 
interest.  The  amount  3|  years  later  was  $  950.  What  was  the 
investment  ? 

26.  The  capital  stock  of  the  Bank  of  France  is  35.6  million 
dollars  less  than  that  of  the  Bank  of  England  and  6.3  million 
greater  than  that  of  the  Imperial  Bank  of  Germany.  The 
combined  capital  stock  of  the  three  banks  is  134.9  million 
dollars.     Find  the  capital  stock  of  each. 

27.  The  capital  stock  of  the  Bank  of  Italy  is  27.7  million 
dollars  less  than  twice  that  of  the  Imperial  Bank  of  Kussia, 
and  the  capital  of  the  Bank  of  Austria-Hungary  is  13.6  million 
greater  than  that  of  the  Bank  of  Russia.  Their  combined 
capital  is  99.1  million  dollars.     Find  the  capital  of  each  bank. 


136  THE  SOLUTION  OF  PROBLEMS 

28.  In  a  bicycle  race  A  starts  32  minutes  ahead  of  B. 
B  rides  at  the  rate  of  20^  miles  per  hour,  while  A  rides  18| 
miles  per  hour.  How  many  miles  from  the  starting  point 
does  B  overtake  A? 

29.  A  squadron  of  warships  sails  13  knots  per  hour.  A 
torpedo  boat  making  27f  knots  leaves  port  19  hours  later  to 
overtake  the  squadron.  In  how  many  hours  after  leaving  port 
will  the  torpedo  boat  overtake  it  ? 

30.  A  man  takes  out  a  life  insurance  policy  for  which  he 
pays  in  a  single  payment.  Thirteen  years  later  he  dies  and  the 
company  pays  $  12,600  to  his  estate.  It  was  found  that  his 
investment  yielded  2  °J0  simple  interest.  How  much  did  he 
pay  for  the  policy  ? 

31.  A  merchant  bought  goods  for  $600  and  some  months 
later  sold  them  for  $648,  making  a  profit  of  2%  per  month. 
How  many  months  elapsed  between  the  purchase  and  the 
sale? 

32.  In  a  building  there  are  at  work  18  carpenters,  7  plumbers, 
13  plasterers,  and  6  hod  carriers.  Each  plasterer  gets  $  1.90 
per  day  more  than  the  hod  carriers,  the  carpenters  get  35 
cents  per  day  more  than  the  plasterers,  and  the  plumbers 
50  cents  per  day  more  than  the  carpenters.  If  one  day's 
wages  of  all  the  men  amount  to  $183.45,  how  much  does  each 
get  per  day  ? 

33.  A  train  running  46  miles  per  hour  leaves  Chicago  for 
New  York  at  7  a.m.  Another  train  running  56  miles  per  hour 
leaves  at  9.30  a.m.  Find  when  the  trains  will  be  15  miles 
apart.     (Two  answers.) 

34.  Divide  the  number  280  into  two  parts  so  that  \  of  one 
part  is  48  less  than  \  of  the  other. 

35.  A  merchant  bought  goods  and  sold  them  5  months  later 
for  $  2687.50,  making  a  gain  on  his  investment  of  1\  %  per 
month.     How  much  did  he  pay  for  the  goods  ? 


REVIEW  PROBLEMS  137 

36.  A  bicyclist  starts  out  riding  12  miles  per  hour,  and  is 
followed  40  minutes  later  by  another  riding  16  miles  per  hour. 
Find  when  they  will  be  5  miles  apart.     (Two  answers.) 

37.  A  father  invested  $  1000  at  6^  %  interest.  When  the 
principal  and  simple  interest  amounted  to  $  2235,  it  was  given 
to  the  son  for  the  expenses  of  his  college  education.  How 
long  had  the  money  been  invested  ? 

38.  Two  trains  start  west  at  the  same  time,  one  from  New 
York  and  the  other  from  Philadelphia.  If  the  New  York  train 
runs  55  miles  per  hour  and  the  Philadelphia  train  47  miles  per 
hour,  how  long  before  they  are  15  miles  apart,  the  distance 
from  New  York  to  Philadelphia  being  90  miles? 

39.  A  man  bought  a  tract  of  coal  land  and  sold  it  a  month 
later  for  $  93,840.  If  his  gain  was  at  the  rate  of  24  %  per 
annum,  what  did  he  pay  for  the  land  ? 

40.  There  is  a  rectangle  whose  length  is  60  feet  more, 
and  whose  width  is  20  feet  less,  than  the  side  of  a  square 
of  equal  area.  Find  the  dimensions  of  the  square  and  the 
rectangle. 

41.  A  number  is  composed  of  two  digits  whose  sum  is  14. 
If  the  digits  are  interchanged,  the  number  is  decreased  by  18. 
What  is  the  number  ? 

42.  What  time  between  7  and  8  o'clock  are  the  hour  and  the 
minute  hands  of  a  clock  together  ? 

43.  Change  104  degrees  Fahrenheit  (fever  heat)  to  Centi- 
grade reading. 

44.  A  steamer  leaves  Liverpool  for  New  York  Saturday, 
9  a.m.,  averaging  18  knots  per  hour.  Seven  hours  later  an- 
other steamer  leaves  New  York  for  Liverpool,  making  20£ 
knots  per  hour.  What  time  (Liverpool  time)  will  the  steamers 
meet  if  the  trans-Atlantic  distance  by  their  course  is  2940 
knots  ? 


CHAPTER   V 
INTRODUCTION   TO   SIMULTANEOUS   EQUATIONS 

104.    Graphic  Representation  of  Statistics. 

A  graphic  representation  of  the  temperatures  recorded  by  the 
U.  S.  Weather  Bureau  at  Chicago  on  December  6  and  7,  1906, 
is  shown  on  the  next  page.     The  readings  were  as  follows : 


3  P.M. 

29° 

9  P.M. 

21° 

3  A.M. 

12° 

9  A.M.      12° 

4  P.M. 

29° 

10  P.M. 

20° 

4  A.M. 

11° 

10  A.M.      13° 

5  P.M. 

28° 

11  P.M. 

17° 

5  A.M. 

10° 

11  A.M.       16° 

6  P.M. 

26° 

12  m't. 

16° 

6  A.M. 

10° 

12  Noon  17° 

7  P.M. 

24° 

1  A.M. 

14° 

7  A.M. 

10° 

1  P.M.      18° 

8  P.M. 

22° 

2  A.M. 

12° 

8  A.M. 

10° 

2  p.m.     20° 

In  the  graph  each  heavy  dot  represents  the  temperature  at  a  cer- 
tain hour.  The  distance  of  a  dot  to  the  right  of  the  heavy  vertical 
line  indicates  the  hour  of  the  day  counted  from  noon  December  6th, 
and  its  distance  above  the  heavy  horizontal  line  indicates  the  ther- 
mometer reading  at  that  hour.  The  lines  joining  these  dots  complete 
the  picture  representing  the  gradual  changes  of  temperature  from 
hour  to  hour. 

Graphs  of  this  kind  are  used  in  commercial  houses  to  represent 
variations  of  sales,  fluctuations  of  prices,  etc.  They  are  used  by 
architects  and  engineers  to  show  the  comparative  strength  of  mate- 
rials, stresses  to  which  they  are  subjected,  and  to  exhibit  multitudes 
of  other  data.  They  are  used  by  the  historian  to  represent  changes 
in  population,  fluctuations  in  mineral  productions,  etc.  In  algebra 
they  are  used  in  solving  many  practical  problems  and  in  helping  to 
understand  many  difficult  processes.     In  the  succeeding  exercises  the 

cross-ruled  paper  is  essential. 

138 


GRAPHING   OF  STATISTICS 


139 


+ 

+ 

+ 

a, 

3 

p 

± 

Ifi 

2 

a 

\ 

+ 

s 

f 

IS 

M. 

I : 

..M 

12 

3  A 

,H 

1 

; 

P.M. 

uTimi 

Line 

EXERCISES 

Make   a  graphic  representation  of   each  of  the   following 
tables  of  data:* 

1.    The  population  of  the  United   States  as   given  by  the 
census  reports  from  1790  to  1900: 

1790  .  .  3.9  (million)  1830 

1800  .  .  4.3  1840 

1810  .  .  7.2  1850 

1820  .  .  9.6  1860 


12.9 

1870  . 

.  38.6 

17.1 

1880  . 

.  50.2 

23.2 

1890  . 

.  62.6 

31.4 

1900  . 

.  76.3 

*In  each  case  the  number  to  be  represented  by  one  space  on  the  cross-ruled 
paper  should  be  chosen  so  as  to  make  the  graph  go  conveniently  on  a  sheet. 
Thus  in  (1)  let  one  small  horizontal  space  represent  two  years  and  one  vertical 


140      INTRODUCTION   TO   SIMULTANEOUS  EQUATIONS 

2.  The  population  of  New  York  city  since  1800 : 

1790  .  .  33  (thousand)  1830  .  .  202  1870  .  .     942 

1800  .  .  60  1840  .  .  312  1880  .  .  1206 

1810  .  .  96  1850  .  .  515  1890  .  .  1530 

1820  .  .  123  1860  .  .  813  1900  .  .  1850 

3.  The  population  of  Chicago  since  1850 : 


1850  . 

.  30  (thousand) 

1870  . 

.  306 

1890 

.  .  1100 

1860  . 

.  109 

1880  . 

503 

1900 

.  .  1698 

4.  The  world's 

yearly  production  of  gold  since  1872 : 

1872  . 

99.6 

(million) 

1883  . 

102.4 

1894 

.  .  181.5 

1873  . 

96.2 

1884  . 

101.7 

1895 

.  .  194.0 

1874  . 

90.7 

1885  . 

108.4 

1896 

.  .  202.3 

1875  . 

97.5 

1886  . 

106.0 

1897 

.  .  236.1 

1876  . 

103.7 

1887  . 

105.8 

1898 

.  .  286.9 

1877  . 

114.0 

1888  . 

110.2 

1899 

.  .  306.7 

1878  . 

119.0 

1889  . 

123.5 

1900 

.  .  254.6 

1879  . 

109.0 

1890  . 

118.9 

1901 

.  .  262.5 

1880  . 

106.5 

1891  . 

130.7 

1902 

.  296.0 

1881  . 

103.0 

1892  . 

146.3 

1903 

.  325.5 

1882  . 

102.0 

1893  . 

157.2 

1904 

.  346.8 

5.  Thei 

world's 

yearly  productic 

n  of  silver 

since 

1872: 

1872  . 

65.0 

(million) 

1883  .  . 

115.3 

1894 

.  214.5 

1873  . 

81.8 

1884  . 

105.5 

1895 

.  208.0 

1874  . 

71.5 

1885  . 

118.5 

1896 

.  203.0 

1875  . 

80.5 

1886  . 

120.6 

1897 

.  207.0 

1876  . 

87.6 

1887  . 

124.3 

1898 

.  218.6 

1877  . 

81.0 

1888  . 

140.7 

1899 

.  217.6 

1878  . 

95.0 

1889  . 

155.4 

1900 

.  224.0 

1879  . 

96.0 

1890  . 

163.0 

1901 

.  223.7 

1880  . 

96.7 

1891  . 

177.0 

1902 

.  208.6 

1881  . 

102.0 

1892  . 

197.7 

1903 

.  220.4 

1882  . 

111.8 

1893  . 

209.1 

1904 

.  217.7 

space  a  million  of  population ;  and  in  (3)  let  one  horizontal  space  represent 
one  year  and  one  large  vertical  space  one  hundred  thousand  of  population. 


GRAPHING   OF  STATISTICS 


141 


6.    Temperature,  at  Port  Conger,  New  York,  and  Singapore. 
The  average  temperature  for  each  half  month  is  given. 


Poet 

New 

Singa- 

Port 

New 

Singa- 

Month 

Conger 

YOKK 

pore 

Month 

Conger 

York 

pore 

Jan.     1-15 

-35° 

30° 

80° 

July 

1-15 

+  32° 

66° 

86° 

16-31 

-40° 

28° 

82° 

16-31 

+  37° 

68° 

86° 

Feb.     1-15 

-45° 

32° 

84° 

Aug. 

1-15 

+  37° 

68° 

87° 

16-28 

-40° 

35° 

84° 

16-31 

+  34° 

66° 

86° 

Mar.    1-15 

-35° 

38° 

85° 

Sept. 

1-15 

+  27° 

62° 

85° 

16-31 

-30° 

40° 

85° 

16-30 

+  20° 

60° 

85° 

April   1-15 

-20° 

45° 

85° 

Oct. 

1-15 

+  8° 

55° 

85° 

16-30 

-10° 

48° 

85° 

16-31 

-  2° 

50° 

84° 

May     1-15 

0° 

50° 

85° 

Nov. 

1-15 

-15° 

48° 

83° 

16-31 

+   8° 

56° 

86° 

16-30 

-20° 

45° 

82° 

June    1-15 

+  16° 

60° 

86° 

Dec. 

1-15 

-28° 

42° 

82° 

16-30 

+  20° 

65° 

86° 

16-31 

-32° 

40° 

81° 

When  the  temperature   is  below  zero,  the  distance  is  measured 
downward  from  the  heavy  horizontal  line,  as  in  the  figure,  page  149. 

7.   Average  monthly  rainfall  at  San  Francisco,  Valparaiso, 
Chili,  and  Quebec : 


(a)  San  Francisco 

(b)  Valparaiso 

(c)  Qttebec 

January    . 
February  . 
March   .     . 

.     5.5  (inches) 
.    4.5 
.     3.5 

0.2 
0.3 

1 

(inches) 

3.2  (inches) 

6.4 

4.4 

April     .    . 
May       .    . 
June      .     . 

.     2.5 
.     1.5 
.     0.5 

2 
3 
4.2 

6.6 
5.1 
2.5 

July       .     . 
Aiagust  .     . 
September 
October 

.    0.2 
.     0.2 
.     0.2 
.     1.5 

2.9 
1.8 
1.8 
0.5 

1.4 

1.8 
1.8 
0.5 

November  . 

.     3.5 

0.4 

0.4 

December  . 

.     5.5 

0.2 

0.2 

Use  10  small  vertical  spaces  for  1  inch  of  rainfall  and  five  horizon- 
tal spaces  for  1  month. 


142      INTRODUCTION  TO   SIMULTANEOUS  EQUATIONS 

8.  After  making  the  graphs  in  exercise  6,  each  on  a  separate 
sheet,  put  all  three  on  the  same  sheet,  using  a  different  colored 
ink  or  pencil  for  each.  In  this  way  the  relative  average  tem- 
peratures are  simultaneously  pictured. 

9.  After  graphing  (a),  (6),  and  (c)  of  example  7,  each  on  a 
separate  sheet,  combine  all  three,  using  different  colors,  as 
in  8. 

10.  Observe  the  weather  reports  in  a  daily  paper  and  make 
a  graph  representing  the  hourly  change  of  temperature  for 
twenty-four  hours. 

GRAPHIC  REPRESENTATION  OF  MOTION 

105.  A  useful  picture  of  the  distance  traversed  by  a  moving 
body  can  be  made  by  a  graph  similar  to  the  preceding. 

E.g.  suppose  a  man  is  walking  3  miles  per  hour.  We  mark  units 
of  time  from  the  starting  point  to  the  right  along  the  horizontal  ref- 
erence line,  and  indicate  miles  traveled  by  the  number  of  units  meas- 
ured vertically  upward  from  this  line.  (See  the  figure  on  the  oppo- 
site page.) 

In  the  figure  each  horizontal  space  represents  1  hour,  and  each  ver- 
tical space  3  miles.  Then  in  1  hour  he  goes  3  miles;  in  5  hours, 
15  miles ;  in  10  hours,  30  miles ;  etc.  The  dots  representing  the 
distances  are  found  to  lie  on  a  straight  line. 

The  graph  shows  at  a  glance  the  answers  to  such  questions  as : 
How  many  miles  does  he  travel  in  4  hours?  in  13  hours?  How  long 
does  it  take  him  to  go  18  miles?  23  miles? 

Again,  suppose  24  hours  later  a  second  man  starts  out  on  a  bicycle 
to  overtake  the  first  man,  and  travels  9  miles  an  hour.  The  line 
drawn  from  the  24-hour  point  shows  the  distance  the  wheelman  trav- 
els in  any  number  of  hours  counting  from  his  time  of  starting.  The 
points  marked  in  this  line  show  how  far  he  has  gone  in  1,  2,  3,  4,  5,  6 
hours,  etc.,  namely,  9,  18,  27,  36,  45,  54,  etc. 

The  point  where  these  two  lines  intersect  shows  in  how  many 
hours  after  starting  the  pedestrian  is  overtaken  and  also  how  far  he 
has  gone. 


GRAPHIC  SOLUTION  OF  PROBLEMS 


143 


i-<o                            ^  r  <?    ■ 

i£       jZ                                         J 

-,** 

.?                               j 

In  like  manner  solve  the  following  by  means  of  graphs,  and 
in  each  case  suggest  other  questions  which  may  be  answered 
from  the  graph : 

1.  In  a  century  bicycle  race  A  averages  17  miles  per  hour; 
B,  who  starts  20  minutes  later,  averages  19  miles  per  hour.  In 
how  many  hours  will  B  overtake  A  ?  Who  will  win  the  race 
and  where  will  the  loser  be  when  the  winner  finishes  ? 

2.  A  starts  for  a  town  12  miles  distant,  walking  3  miles  per 
hour,  li  hours  later  B  starts  for  the  same  place,  driving  1\ 
miles  per  hour.  When  does  B  overtake  A  ?  Where  is  A  when 
B  reaches  town  ? 


144      INTRODUCTION   TO  SIMULTANEOUS  EQUATIONS 

3.  In  a  mile  race  A  runs  6  yards  per  second  and  B  5  yards 
per  second.  B  has  a  start  of  250  yards.  Who  will  win 
the  race  ?     How  far  in  the  lead  is  the  winner  at  the  finish  ? 

Let  1  vertical  small  space  represent  20  yards,  and  1  horizontal  space 
represent  5  seconds. 

106.  Illustrative  Problem.  A  man  rides  a  bicycle  into  the 
country  at  the  rate  of  8  miles  per  hour.  After  riding  a  certain 
distance  the  wheel  breaks  down  and  he  walks  back  at  the  rate 
of  3  miles  per  hour.  How  far  does  he  go  if  he  reaches  home 
11  hours  after  starting  ? 


\J 

.•■>a  &._    _                   >__    __    3^K _            _    _ __    _ 

3-  G                         -f                     -Sw 

:**:£ —    zztzz                         *%       ~              ~ 

\ j         f                                        **». 

In  this  graph  each  large  horizontal  space  represents  1  hour,  and 
each  small  vertical  space  represents  1  mile.  The  problem  is  solved 
as  follows : 

(1)  Construct  the  line  representing  the  outward  journey  at  the 
rate  of  8  miles  per  hour,  extending  this  line  indefinitely. 

(2)  Beginning  at  the  point  corresponding  to  11  hours,  find  the 
points  representing  his  position  at  each  pi-eceeding  hour.  The  line 
connecting  these  points  represents  the  homeward  journey  at  the  rate 
of  3  miles  an  hour.  Extend  this  line  until  it  meets  the  first  line. 
The  point  where  the  lines  meet  represents  3  hours  and  24  miles, 
which  is  the  answer  required  in  the  problem. 


GRAPHIC  SOLUTION   OF  PROBLEMS  145 

PROBLEMS 

Solve  the  following  problems  by  means  of  graphs.  In  each 
case  prepare  a  list  of  questions  which  may  be  answered  from 
the  graph. 

1.  A  man  rows  18  miles  per  hour  down  a  river  and  2  miles 
per  hour  returning.  How  far  down  the  river  can  he  go  if  he 
wishes  to  return  in  10  hours  ? 

2.  A  man  goes  from  Chicago  to  Milwaukee  on  a  train  run- 
ning 42^  miles  per  hour,  and  returns  immediately  on  a  steamer 
going  17  miles  per  hour.  Find  the  distance,  if  the  round  trip 
requires  7  hours. 

3.  A  pleasure  trip  from  New  York  to  Atlanta  by  steamer 
and  return  by  rail  occupied  77  hours.  Find  the  distance,  if 
the  rate  going  was  16  miles  per  hour  and  returning  40  miles 
per  hour. 

Let  one  small  horizontal  space  represent  one  hour  and  one  small 
vertical  space  16  miles. 

4.  A  invests  $1000  at  5%  and  B  invests  $5000  at  4%. 
In  how  many  years  will  the  amount  (principal  and  interest)  of 
A's  investment  equal  the  interest  on  B's  investment? 

Let  one  large  horizontal  space  represent  one  year  and  one  small 
vertical  space  $50.  Then  the  line  representing  ^4's  amount  starts  at 
the  point  marked  $1000,  and  rises  one  small  vertical  space  each  year. 
The  line  representing  B'a  interest  starts  at  the  zero  point  and  rises 
four  small  vertical  spaces  each  year. 

5.  In  how  many  years  will  the  interest  on  $  6000  equal  the 
amount  on  $  2000  if  both  are  invested  at  5  %  ? 

6.  A  invests  $500  at  6%  and  B  invests  $  1000  at  5%.  In 
how  many  years  will  A's  interest  differ  by  $300  from  .B's  ? 

Let  one  small  vertical  space  represent  $20.  In  this  case  both  lines 
start  from  the  zero  point.  Find  the  point  on  one  line  which  is  three 
large  spaces  vertically  above  the  corresponding  point  on  the  other 
line. 


146      INTRODUCTION  TO  SIMULTANEOUS  EQUATIONS 

7.  Construct  a  graph  representing  the  relation  between  the 
Fahrenheit  and  Centigrade  thermometer  readings. 

Let  the  hue  at  the  bottom  of  the  sheet  and  the  vertical  line  four 
large  spaces  from  the  left  margin  be  the  reference  lines.  Let  one 
small  horizontal  space  represent  a  degree  C,  and  one  smal]  vertical 
space  a  degree  F. 

From  F  =  32  +  |  C  (page  129),  we  find  that  if  C  =  0°  F  =  32°,  if 
C  =  10°  F=50°,  if  C  =  20°  F  =  68°,  if  C  =  30°  F  =  86°,  etc.  Mark 
the  point  representing  each  pair  of  readings,  and  draw  the  straight 
line  connecting  these  points.     This  is  the  required  graph. 

8.  From  this  graph  read  the  answers  to  the  following 
questions : 

Find  C  when  F=  41°,  F  =  59°,  F=  79°,  F  =  14°. 
Find  F  when  C=  35°,  C  =  40°,  C=  -  5°,  C=  -  15°. 

From  the  graph  it  is  possible  to  find  any  Fahrenheit  reading  when 
the  corresponding  Centigrade  is  given,  and  also  to  find  any  Centigrade 
reading  when  the  corresponding  Fahrenheit  is  given.  Thus  the  graph 
shows  to  the  eye  all  the  information  contained  in  the  equation  F  =  32 
+ 1 C.  In  what  follows  we  shall  consider  in  detail  how  to  represent 
equations  in  this  manner. 

GRAPHIC  REPRESENTATION  OF  EQUATIONS 
107.  In  all  the  graphs  thus  far  constructed  two  lines  at  right 
angles  to  each  other  have  been  used  as  reference  lines.  These 
lines  are  called  axes.  The  location  of  a  point  in  the  plane  of 
such  a  pair  of  axes  is  completely  described  by  giving  its  distance 
and  direction  from  each  of  the  axes.  The  direction  to  the 
right  of  the  vertical  axis  is  denoted  by  a  positive  sign,  and  to 
the  left,  by  a  negative  sign j  while  direction  upward  from  the 
horizontal  axis  is  positive,  and  downward,  negative. 

The  horizontal  line  is  usually  called  the  jr-axis  and  the  ver- 
tical line  the  /-axis.  The  perpendicular  distance  of  any  point 
P  from  the  ?/-axis  is  called  the  abscissa  of  the  point,  and  its  dis- 
tance from  the  x-axis  is  called  its  ordinate.  The  abscissa  and 
ordinate  of  a  point  are  together  called  its  coordinates. 


GRAPHS  REPRESENTING  EQUATIONS 


147 


+ 

. 

1 

> 

=r> 

<W 

7J 

P 

( 

8^ 

M 

+ 

< 

•f 

_H 

i 

_ 

3 

_ 

) 

1 

* 

l 

, 

+  :; 

t 

i 

I 

C 

Zx 

)f 

X 

j 

1 

— 

, 

-• 

I 

- 

J 

I 

*; 

E.g.  the  abscissa  of  point  P  in  the  above  figure  is  3  and  its 
ordinate  2,  or  we  may  say  the  coordinates  of  P  are  3  and  2,  and  indi- 
cate it  thus :  P  :  (3,  2),  writing  the  abscissa  first.  In  like  manner  for 
the  other  points  we  write  Q:  (-  1,  3),  R:  (-  2,  0),  S :  (-  3,  -  4), 
and  T  :  (2,  —  3).  (For  convenience  upper  signs  are  used  in  the 
figure.) 

We  see  that  in  this  manner  every  point  in  the  plane  corre- 
sponds to  a  pair  of  numbers  and  that  every  pair  of  numbers  cor- 
responds to  a  point.  This  scheme  of  locating  points  by  two 
reference  lines  is  already  familiar  to  the  pupil  in  geography, 
where  cities  are  located  by  latitude  and  longitude,  that  is,  by 
degrees  north  or  south  of  the  equator  and  east  or  west  of  the 
meridian  of  Greenwich. 


148      INTRODUCTION  TO  SIMULTANEOUS  EQUATIONS 

EXERCISES 

1.  With  any  convenient  scale,  locate  the  following  points  : 
(2,6),  (-3,5),  (0,1),  (1,0),  (0,0),  (0,-1),  (0,-5), 
(-5,0),   (2i,  5ft,    (-4,-8),    (3,-10),    (-10,3). 

2.  Calling  north  + ,  south  — ,  east  + ,  west  — ,  so  that 
(—  5°,  8°)  means  the  point  on  a  map  whose  longitude  is  5° 
west  and  whose  latitude  is  8°  north,  verify  the  following  on  a 
map:  New  York  (-  73°  57',  40°  48'),  Chicago  (-87°  35',  41° 
51'),  Peking  (116°  30',  39°  50'),  Sydney,  New  South  Wales, 
(151°  15',  -33°  51'),  Santiago,  Chili,  (-  70°  39',  -  33°  25'). 

3.  On  a  map  of  South  America  give  approximately  the  lo- 
cation of  the  following  cities,  using  the  notation  of  the  pre- 
ceding exercise :  Caracas,  Rio  de  Janeiro,  Bogota,  Valparaiso, 
Lima,  and  Panama. 

4.  On  a  map  of  Africa  locate  in  similar  manner  the  cities  or 
islands  situated  at  the  following  points:  (— 5°  40',  —  16°), 
(-  14°  30',  -  8°),  (30°  18',  30°  1'),  (-  5°  40',  36°),  (40°  40', 
-  15°),  (18°  20',  -  33°  55'). 

5.  Locate  the  following  series  of  points  and  then  see  if  a 
straight  line  can  be  drawn  through  them :  (0,  0),  (1,  1),  (2,  2), 
(3,  3),  (4,  4),  (-  1,  -  1),  (-  2,  -  2),  (-  3,  -  3).  Name  still 
other  points  lying  on  the  same  line. 

6.  Locate  the  following  and  connect  them  by  a  line:  (1,0), 
(1,  2),  (1,  3),  (1, 4),  (1,  5),  (1,  -  2),  (1,  -  3),  (1,  -  4),  (1,  -  5). 
Name  other  points  in  this  line. 

7.  Draw  the  line  every  one  of  whose  points  has  its  hori- 
zontal distance  —  2,  also  the  line  every  one  of  whose  points  has 
its  vertical  distance  +  3. 

8.  Locate  the  following  points  and  see  if  a  straight  line  can 
be  passed  through  them :  (1,  0),  (0,  1),  (2,  -  1),  (3,  -  2), 
(4-3),  (-1+2),  (-2,  3),  (-3,  4),  (-4,  5),  &  *),  &  f),  (f,  *). 
Can  you  name  other  points  on  this  line  ? 


GRAPHS  REPRESENTING  EQUATIONS  149 


Temperature  chart  for  Port  Conger.  (See  page  141.) 


150      INTRODUCTION  TO  SIMULTANEOUS  EQUATIONS 


108.  In  the  preceding  exercises,  in  certain  cases,  a  series 
of  points  has  been  found  to  lie  on  a  straight  line  as  in 
examples  6,  7,  and  8.  Evidently  this  could  not  happen 
unless  the  points  were  located  according  to  some  definite 
scheme  or  law. 

Illustrative  Problem.  Locate  a  series  of  points  whose  co- 
ordinates are  values  of  x  and  y  which  satisfy  :  3  x  +  4  y  =  12. 

We  see  that  x  =  0,  y  =  3,  also  x  =  4,  y  =  0  are  pairs  of  such  values. 
Evidently  as  many  pairs  of  values  as  we  please  may  be  found  by 
giving  any  value  to  x  and  then  solving  the  equation  to  find  the  corre- 
sponding value  of  y.     A  table  may  thus  be  constructed  as  follows : 

Let  x  =  0,  4,      8,     12,  -4,-8,  etc. 

Then  y  -  3,  0,-3,-6,       6,      9,  etc. 


-i 

n 

-t- 

- 

' 

4 

4 

+ 

* 

+ 

+ 

f 

-. 

T 

-( 

It 

h| 

"■ 

n 

— 

_ 

p 

^ 

' 

These  pairs  of  values  for  x  and  y  correspond  to  the  points  as 
plotted  in  the  figure,  and  they  are  found  to  lie  on  a  straight 
line.     This  line  is  called  the  graph  of  the  equation. 


GRAPHS  REPRESENTING  EQUATIONS  151 

Let  the  student  find  other  pairs  of  numbers  which  satisfy  this 
equation  and  see  if  the  corresponding  points  lie  on  this  line.  Also 
find  the  numbers  which  correspond  to  any  chosen  point  on  this  line 
and  see  whether  they  satisfy  the  equation. 

109.  Since  an  equation  like  3  x  -f  4y  =  12  is  satisfied  by  in- 
definitely many  pairs  of  values  of  x  and  y,  it  is  common  to  call 
the  unknowns  in  such  an  equation  variables.  Indeed  if  a  point 
be  thought  of  as  moving  along  the  graph  of  this  equation  the 
values  of  x  and  y  corresponding  to  the  moving  point  continually 
vary,  but  always  so  that  3  x  +  4y  =  12. 

110.  Definitions.  An  equation  is  said  to  be  of  the  first  degree 
in  x  and  y  if  it  contains  each  of  these  letters  in  such  a  way  that 
neither  x  nor  y  is  multiplied  by  itself  or  by  the  other. 

E.g.  13  x  —  5  y  =  14  is  of  the  first  degree,  while  2  xy  —  x  =  5  and  3  x 
—  5  y2  =  13  are  not  of  the  first  degree  in  x  and  y. 

Every  equation  of  the  first  degree  in  two  variables  has  for 
its  graph  a  straight  line ;  hence  such  an  equation  is  commonly 
called  a  linear  equation. 

111.  To  graph  an  equation  of  the  first  degree  it  is  only 
necessary  to  find  two  points  on  the  graph  and  draw  a  straight 
line  through  them. 

E.g.  In  graphing  the  equation  x  —  y  =  5,  we  choose  x  =  0  and  find 
y  =  —  5,  and  choose  y  =  0  and  find  x  =  5  and  plot  the  points  (0,  —  5) 
and  (5,  0).     The  line  through  these  points  is  the  one  required. 

EXERCISES 

Construct  the  graph  for  each  of  the  following  equations : 

1.  3x  +  2y  =  l.        5.   5-2y=12.  9.   3a-4?/=-7. 

2.  5x-3y=-3.     6.   3x  +  5y=  — 15.    10.   3x-4y=-12. 

3.  7x  +  10y=2.       7.   2x-y  =  0.  11.    7y  =  9»— 63. 

4.  x  +  2y  =  0.  8.   3x-4y  =  7.  12.    x  =  5y  +  3. 


152      INTRODUCTION   TO   SIMULTANEOUS  EQUATIONS 

112.    Illustrative  Problem.    Graph  on  the  same  axes  the  two 
equations  x  +  y  =  4  and  y  —  x  =  2. 


/ 

/ 

/ 

/ 

t 

/ 

f 

Lt 

i 

^ 

' 

h 

1 

1 

-1 

t 

- 

. 

- 

n 

+ 

t 

2 

+ 

:i 

-1 

i 

~J 

1 

— 

Solution.  The  two  graphs  are  found  to  intersect  in  the  point  (1,  3). 
Since  the  point  lies  on  both  lines,  its  coordinates  should  satisfy  both 
equations,  as  indeed  they  do.  Since  these  lines  have  only  one  point 
in  common,  there  is  no  other  pair  of  numbers  which,  when  substituted 
for  the  variables  x  and  y,  can  satisfy  both  equations. 

Hence  x  =  1,  y  =  3,  which  is  written  (1, 3),  is  called  the  solu- 
tion of  this  pair  of  equations. 

113.  Definition.  These  two  equations  are  called  independ- 
ent because  their  graphs  are  distinct.  They  are  called  simulta- 
neous because  there  is  at  least  one  pair  of  values  of  x  and  y 
which  satisfy  both. 


ELIMINATION  BY  SUBSTITUTION  153 

114.  Since  two  straight  lines  intersect  in  but  one  point,  it 
follows  that  two  linear  equations  which  are  independent  and 
simultaneous  have  one  and  only  one  solution. 

EXERCISES 

Graph  the  following  and  thus  find  the  solution  of  each  pair 
of  equations : 

L    f2x-3y  =  25,  6     ty  +  3x  =  7, 

Lie  -f-  3/  =  5.  \2y-\-x=  —6. 

2    |5*  +  6y  =  7,  ?     \x-2y  =  2, 

\2x-y  =  -A.  '    V2x-y=-2. 

g     f5*+3y»l,  8     f5x-7y  =  21, 

i2x  +  y  =  — 4.  (x  —  £y=—  1. 

4  J6a;  +  8y  =  16,  Q     \5x  +  2y  =  8, 
\2x-3y  =  ll.  '    (2x-3y=-12. 

5  \'3x-4:y  =  l,  1Q     tx  +  3y=-6, 
\2x-7y  =  5.  '    \2x-±y=-12. 

SOLUTION   OF   SIMULTANEOUS   EQUATIONS   BY   SUBSTITUTION 

115.  A  pair  of  linear  equations  may  be  solved  without  con- 
structing their  graph. 

Illustrative  Problem.  The  sum  of  'two  numbers  is  35  and 
their  difference  is  5.     What  are  the  numbers  ? 

Solution.  Let  x  =  the  greater  number  and  y  the  smaller. 

Then  fz  +  y  =  35,  (1) 

and  1  x  —  y  =  5.  (2) 

From  (1)  by  5,  y  =  35  -  x.  •  (3) 

Substituting  35  —  x  for  y  in  (2),   x  —  (35  —  x)  =  5.  (4) 

Solving,  x  =  20.  (5) 

Substituting  x  =  20  in  (1),  y  =  15.  (6) 


154      INTRODUCTION  TO  SIMULTANEOUS  EQUATIONS 

116.  In  most  problems  solved  heretofore  there  have  been  two 
or  more  unknown  numbers.  In  forming  the  equations  the 
object  has  been  to  express  all  but  one  of  the  unknowns  in 
terms  of  that  one.  In  the  case  of  two  unknowns  this  is  now 
done  more  systematically  as  follows : 

(a)  State  two  equations  involving  the  two  unknowns,  as  (1) 
and  (2)  above. 

(6)  Solve  one  of  these  equations  for  one  unknown  in  terms 
of  the  other  as  in  (3)  above. 

(c)  Substitute  in  the  other  equation  the  value  of  this  un- 
known as  thus  expressed,  obtaining  as  in  (4)  one  equation  in 
one  unknown. 

This  equation  (4)  is  the  same  as  equation  (1)  on  page  111, 
where  this  problem  was  solved  by  means  of  one  unknown. 
The  solution  given  here  amounts  to  a  formal  tabulation  of  the 
steps  there  taken  in  obtaining  the  equation,  g—(35—g)  =  5. 

Since  equation  (4)  contains  but  one  of  the  unknowns,  the 
other  is  said  to  be  eliminated. 

The  process  here  used  is  called  elimination  by  substitution. 

Illustrative  Problem.    Solve  the  equations : 


\2x  +  3y  =  13. 
\5x-6y=-8. 

(1) 

(2) 

From  (1)  by 

S  and  D, 

13-2* 

V=        3       * 

(3) 

Substituting 

in  (2) 

5x      6Q3-2*)=      8. 
3 

(4) 

ByF,V, 

5  x- 2(13  -  2x)=-8. 

(5) 

By  IV,  VII, 

5x-26+  ix  =  -k 

(6) 

By  I,  A, 

9  x  =  18. 

(7) 

By  A 

x  =  2. 

(8) 

From  (3), 

13  -  2  •  2       - 

y  =        .,        = a- 

(9) 

Verify  this  by  drawing  the  graphs  and  also  by  substituting  these 
values  of  x  and  y  in  (1)  and  (2). 


ELIMINATION  BY  SUBSTITUTION 


155 


EXERCISES 

Solve  the  following  pairs  of  linear  equations  by  eliminating 
one  of  the  variables  by  substitution,  and  check  by  substituting 
the  results  in  the  original  equations. 


i. 


2. 


4. 


5. 


7. 


8. 


9. 


10. 


11. 


x  —  y  =  10. 

x-y  =  -3, 
a;  +  4  y  =  12. 

2x  +  3y  =  5, 
7x-5y  =  33. 

3  a;  -  4  y  =  8, 

4  a;  +  3  y  =  -  6. 

2  x  -  4  y  =  8, 

3  a?  +  2  y  =  4. 

a;  +  2y  =  4, 

2  a;  +  y  =  -  1. 

3  x  —  4  y  =  8, 
2aj  +  3y  =  ll. 

5  a;  +  9  y  =  19, 

3  x  —  y  =  5. 

4  y  -  2  x  =  3, 

2  y  +  5  a;  =  6. 

3  x  -  7  y  =  -  11, 
2  a;  +  y  =  4. 

5<c  — 3y  =  4— 2a;+7y, 

5  y  +  a;  =  7. 


12. 


13. 


14. 


15. 


16. 


17. 


18. 


19. 


20. 


21. 


22. 


|3y  +  5a;  =  12  +  2a;, 
tl7a;-y  =  4y-20. 

f6y-a;  =  7  +  4y, 
\5x  +  8y  =  l. 

(6  +  x  +  y  =  2x-l, 
l3y  +  z  =  6y  +  9. 

x-y  =  37, 

+  3y  =  314  +  13y. 

2z-3y  =  y  +  6, 
a?  +  2y  =  4y  +  3. 

y  +  5a;  =  2a;  +  5, 
y  -  3  x  =  19. 


\2x 


l2y- 

J5a;  +  3y  =  0, 
I2aj  +  y  =  l. 

2x  +  3y=6x-l, 


|2a;  +  3y  = 
I3a;-2v  = 

r 


y  =  3. 
5  x  -  3  y  =  0, 


2a;  +  2-6y  =  2-a;. 

|6a;+2y=23, 
Il0aj-5v=21. 

{ 


,y: 

3a-7y=15, 
5aj  +  4y  =  11. 


Solve  21  and  22  by  means  of  graphs  and  also  by  elimination 
by  substitution.  Notice  that  the  latter  method  gives  a  more 
accurate  solution  than  can  be  obtained  from  the  graph. 


156      INTRODUCTION   TO   SIMULTANEOUS  EQUATIONS 

SOLUTION   OF   SIMULTANEOUS   EQUATIONS   BY   ADDITION    OR 
SUBTRACTION 

117.  Illustrative  Examples.     Solve 

<x  +  2y  =  7,  (1) 

(Sx-2y  =  5.  (2) 

Adding  the  members  of  these  equations,  +  2  y  and  —  2  y  cancel. 

Hence,                                          4x=12,  (3) 

x  =  3.  (4) 

Substituting  in  (1),         3  +  2  y  =  7,  (5) 

y=2.  (6) 
Verify  this  by  drawing  the  graphs,  and  also  by  substituting  x  =  3, 
y  =  2,  in  (1)  and  (2). 

If  one  of  the  variables  does  not  already  have  the  same  co- 
efficient in  both  equations,  the  solution  may  be  obtained  as 
follows : 


Solve  the  equations 


[7a?-f3y  =  4-y  +  4a>,  (1) 

L3cc  — 2/  =  4?/  — 2  — x.  (2) 


From  (1)  by  A,  S,             3x  +  iy  =  4.  (3) 

From  (2)  by  A,  S,              4x-5y  =  -2.  (4) 

From  (3)  by  M,              12  x  +  16  y  =  16.  (5) 

From  (4)  by  M,              12x-15y  =  -Q.  (6) 

Subtracting  (6)  from  (5),         31  y  =  22.  (7) 

From  (7)  by  D,                                y  =  f  { .  (8) 

Substituting  in  (3),                          x  =  \\.  (9) 
Hence,  the  solution  is  Jf,  f|, 

118.  The  process  used  in  the  solution  just  given  is  called 
elimination  by  addition  or  subtraction. 

This  method  is  usually  simpler  than  elimination  by  substitu- 
tion, since  the  latter  frequently  involves  fractions. 


ELIMINATION  BY  ADDITION  OR   SUBTRACTION     157 
EXERCISES 

Solve  the  following  pairs  of  equations  by  addition  or  sub- 
traction. Substitute  the  results  in  the  given  equations  in  each 
case  to  test  the  accuracy  of  the  solution. 

L    j2a;  +  3?/  =  22,  6      i5x+10y=-7, 

\x  —  y  =  l.  \2x+5y  =  —  2. 

2  \5x-2y  =  2l,  ?>    l5x  +  3y=-2, 
\x  —  y  =  6.  \3x  +  2y=  —  1. 

3  iGx+30  =  8y,  8     |3a  +  7&  =  7, 
l3y  +  17=2-3a;.  l5a  +  3&  =  29. 

4  1 8  x  —  4  y  =  12  x,  9      fr  =  3s-19, 
l4a  +  2?/  =  3  +  4y.  U  =  3r-23. 


5. 


tx  +  6y  =  2x-16,  1Q     J2p  =  5g-16, 

13  a -2?/ =  24.  l7g=  —  3i>-|-5. 


Solve  the  following  by  either  process  of  elimination : 

lt    j7m  =  2n-3,  6     |15 &  =  10-20Z, 

ll97i  =  6m  +  89.  l25A:-30Z  =  80. 


2. 


19n  =  6m  +  89 

|6c  +  15d=-6,  ? 

l21d-8c=-74.  1 


6c  +  15d=-6,  7     \28x+Uy  =  23, 

14  #  —  14  y  =  1. 


5. 


I2x-3y  =  4:,  8      i5x  +  2y  =  x  + 18, 
12?/  — 3aj=  —21.  l2a  +  3i/  =  3:B  +  27. 

(u  +  v  =  27,  9     J7?/  — #  =  a;  — 17, 
4v  =  19-|m.  12  y  + 3  a;  =  38. 

|7a  =  l  +  10y,  1Q     |6a;  +  22/=-2, 
ll6?/=10a-l.  \x-4y  =  -35. 


158      INTRODUCTION  TO   SIMULTANEOUS  EQUATIONS 
(3x-y  =  2x-l,  16     J' 


11. 


3x-y  =  2x-l,  16     |7o5  —  42/  =  3, 

+  y  =  14.  I5x  +  8y  =  6. 


12. 


13. 


|2ar  +  32/  =  5,  1?      (12y  -  10a  =  -  6, 


l7a-2y  =  74.  l5.v-9a;  =  l 

{ 


14     f6*/  +  2z  =  ll,  19     {7x  +  4:y  =  3, 


lB     f4y+9x  =  -5,  2Q     f34z  +  702/  =  4, 


3?/ +  12  a  =  18.  12  a;  +  3?/ =  25. 

f  34  x  +  70  ?/  = 
ls  +  y  =  -5.  l5a-8y  =  -36. 

119.  The  equations  thus  far  given  have  for  the  most  part 
been  written  in  a  standard  form,  ax  +  by  =  c,  in  which  all  the 
terms  containing  x  are  collected,  likewise  those  containing  y, 
and  those  which  contain  neither  variable.  When  the  equations 
are  not  given  in  this  form,  they  should  be  so  reduced,  as  in  the 
second  solution  on  page  156,  before  applying  any  method  of 
elimination,  and  also  before  solving  by  means  of  graphs. 

EXERCISES 

After  reducing  each  of  the  following  pairs  of  equations  to 
the  standard  form,  solve  by  graphing,  or  by  means  of  either 
process  of  elimination,  as  seems  best  available. 


1. 


ix-U  =  7y,  3     jr  +  l=-4s, 

Uy  +  l  =  x.  l2s  =  13-5r. 

f     _l-8n 
tl6x-3y  =  7x,  4.      m 5      ' 

Uy  =  7cc  +  5.  l3ra  +  5n  =  l. 


ELIMINATION  BY  ADDITION  OR   SUBTRACTION     159 


s     rm  —  n  =  16, 
l3»  =  8-27i. 


Ta?- 15 


=  2/, 


6.  3 

1.2  a:  — y  =  3 


fa;  — 3 


=  -2, 


7.        5y 

la;  +  7  y  =  6. 

8     |3^-5=-y, 

\8y  +  76  =  5x. 

fa  +  45  =  14, 
l3a-6  =  14. 

10.  2     ^     2  ' 

[2x-y  =  16. 


11.    < 


5     +~3~-8' 
2x_-y_3y-x_ 
2  4  *' 


7m  +  8    7«  — 1 


12. 


2  m—  4     rt  — 1  x 

2      +~3~~~*' 


(x  -  3  |  y  +_ 


13.     < 


a  +  7      2y-4__5 


14. 


12  7 

f8a-3     55-2 

9       +      3 
2a+7     35+10 


13, 


10 


=  -34. 


15. 


Ty-4     2a?-3 

5      "*"      2      ~   ' 
6»-3      2y+l 
5^5 


16.    - 


3y  +  7      5a:-7 


2  3 

2a;-4     2y-l 


10, 


3  4 

f5  +  3p     5g-2 

17.  J      7  4 
Up  +  8?  =  108. 

(3x-2y  =  ±, 

18.  J2te  — 1     7y-4 
I      5  3 


-2i- 


-2, 


=  -19. 


f  5  x  +  7  y  =  89$, 
19.    \2x-4     6y-l 

[     3      +~5~  ~16*- 


20. 


f32a;-9  2/  =  299, 

2jc-5  _  3j/-jL 
7  2 


=  -16. 


160      INTRODUCTION  TO   SIMULTANEOUS  EQUATIONS 

PROBLEMS 

Solve  the  following  problems,  using  two  variables  in  each 
case. 

1.  A  rectangular  field  is  32  rods  longer  than  it  is  wide. 
The  length  of  the  fence  around  it  is  308  rods.  Find  the  di- 
mensions of  the  field. 

2.  Find  two  numbers  such  that  7  times  the  first  plus  four 
times  the  second  equals  37 ;  while  3  times  the  first  plus  9 
times  the  second  equals  45. 

3.  A  certain  sum  of  money  was  invested  at  5%  interest 
and  another  sum  at  6%,  the  two  investments  yielding  $980 
per  annum.  If  the  first  sum  had  been  invested  at  6%  and 
the  second  at  5%,  the  annual  income  would  be  $1000.  Find 
each  sum  invested. 

4.  The  combined  weight  of  3  cubic  centimeters  of  platinum 
and  50  cubic  centimeters  of  poplar  is  84  grams,  and  the  weight 
of  1  cubic  centimeter  of  platinum  and  150  cubic  centimeters  of 
poplar  is  80  grams.  Find  the  weight  of  1  cubic  centimeter 
of  each. 

5.  The  combined  distance  from  the  sun  to  Jupiter  and 
from  the  sun  to  Saturn  is  1369  million  miles.  Saturn  is  403 
million  miles  farther  from  the  sun  than  Jupiter.  Find  the 
distance  from  the  sun  to  each  planet. 

6.  The  sum  of  the  distances  from  the  sun  to  the  fixed 
stars  Altair  and  Capella  is  45.3  light-years.  Twice  the  dis- 
tance of  Altair  plus  3  times  that  of  Capella  is  119.6  light- 
years.     Find  the  distance  from  each  star  to  the  sun. 

7.  Find  two  numbers  such  that  7  times  the  first  plus  9 
times  the  second  equals  116,  and  8  times  the  first  minus  4 
times  the  second  equals  4. 

8.  The  sum  of  two  numbers  is  108.  8  times  one  of  the 
numbers  is  9  greater  than  the  other  number.  Find  the 
numbers.    . 


PROBLEMS  IN  TWO   VARIABLES  161 

9.  Two  investments  of  $24,000  and  $16,000  respectively 
yield  a  combined  income  of  $840.  The  rate  of  interest  on 
the  larger  investment  is  1  %  greater  than  that  on  the  other. 
Find  the  two  rates  of  interest. 

10.  A  father  is  twice  as  old  as  his  son.  Twenty  years 
ago  the  father  was  six  times  as  old  as  his  son.  How  old  is 
each  now? 

11.  If  the  length  of  a  rectangle  is  increased  by  3  feet  and 
its  width  decreased  by  1  foot,  its  area  is  increased  by  3  square 
feet.  If  the  length  is  increased  by  4  feet  and  the  width  de- 
creased by  2  feet,  the  area  is  decreased  by  3  square  feet. 
What  are  the  dimensions  of  the  rectangle  ? 

Let  I  =  the  original  length  and  w  the  width, 

then  ,        (I  +  3)  (w  -  1)  =  ho  +  3,  (1) 

and  (Z  +  4)(w-2)  =  Zw-3.  (2) 

From  (1)  by  XIII,  Iw  +  3  w  -  I  -  3  =  Iw  +  3.  (3) 

From  (2)  by  XIII,  Iw  +  4  w  -  2  I  -  8  =  Iw  -  3.  (4) 

From  (3)  and  (4)  3w-  1  =  6,  (5) 

and  4  w  -  2  I  =  5.  (6) 

(5)  and  (6)  may  now  be  solved  in  the  ordinary  manner. 

12.  The  greatest  distance  from  the  earth  to  Venus  is  160 
million  miles  and  the  shortest  distance  is  26  million  miles. 
How  far  from  the  sun  are  Venus  and 

the  earth,  assuming  that  they  move 
around  the  sun  in  concentric  circles 
with  the  sun  at  the  center  ? 


13.  Two  weights,  35  and  40  pounds 
respectively,  balance  when  resting 
on  a  beam  at  certain  unknown 
distances  from  the  fulcrum.  If 
15    pounds    is    added    to    the    35-lb. 

weight,   the    40-lb.    weight    must    be    moved    2   feet   farther 
from  the  fulcrum  in  order  to  maintain  the  balance.     What 


162      INTRODUCTION  TO   SIMULTANEOUS  EQUATIONS 

was  the  original  distance  from  the  fulcrum  to  each  of  the 
weights  ? 

If  the  distances  from  the  fulcrum  to  the  weights  are  di  and  di  re- 
spectively, then  by  the  formula,  w\dx  =  w2d2,  given  on  page  121, 
we  have  35  dx  =  40  d2  and  50  di  =  40  (d%  +  2). 

14.  A  steamer  on  the  Mississippi  makes  6  miles  per  hour 
going  against  the  current  and  19^  miles  per  hour  going  with 
the  current.  What  is  the  rate  of  the  current  and  at  what 
rate  can  the  steamer  go  in  still  water  ? 

15.  A  man  starts  at  7  a.m.  for  a  walk  in  the  country.  At 
10  a.m.  another  man  starts  on  horseback  to  overtake  the  pedes- 
trian, which  he  does  at  1  p.m.  If  the  rate  of  the  horseman  had 
been  two  miles  per  hour  less,  he  would  have  overtaken  the 
pedestrian  at  4  p.m.     At  what  rate  does  each  travel  ? 

16.  A  camping  party  sends  a  messenger  with  mail  to  the 
nearest  post  office  at  5  a.m.  At  8  a.m.  another  messenger  is 
sent  out  to  overtake  the  first,  which  he  does  in  2\  hours.  If 
the  second  messenger  travels  5  miles  per  hour  faster  than  the 
first,  what  is  the  rate  of  each  ? 

17.  There  are  two  numbers  such  that  3  times  the  greater  is  18 
times  their  difference,  and  4  times  the  smaller  is  4  less  than 
twice  the  sum  of  the  two.     What  are  the  numbers  ? 

18.  Roy  is  three  times  as  old  as  Fred  was  8  years  ago.  Five 
years  from  now  Roy  will  be  16  years  less  than  twice  as  old 
as  Fred.     How  old  is  each  now  ? 

19.  A  picture  is  3  inches  longer  than  it  is  wide.  The  frame  4 
inches  wide  has  an  area  of  360  square  inches.  What  are  the 
dimensions  of  the  picture  ? 

20.  The  difference  between  2  sides  of  a  rectangular  wheat 
field  is  30  rods.  A  farmer  cuts  a  strip  5  rods  wide  around  the 
field,  and  finds  the  area  of  this  strip  to  be  1\  acres.  What 
are  the  dimensions  of  the  field  ? 


PROBLEMS  IN   TWO   VARIABLES  163 

21.  The  sum  of  the  length  and  width  of  a  certain  field  is 
260  rods.  If  20  rods  are  added  to  the  length  and  10  rods 
to  the  width,  the  area  will  be  increased  by  3800  square  rods. 
What  are  the  dimensions  of  the  field  ? 

22.  In  a  number  consisting  of  two  digits  the  sum  of  the 
digits  is  12.  If  the  order  of  the  digits  is  reversed  the  number 
is  decreased  by  36.     What  is  the  number  ? 

23.  A  bird  attempting  to  fly  against  the  wind  is  blown 
backward  at  the  rate  of  1\  miles  per  hour.  Flying  with  a 
wind  \  as  strong,  the  bird  makes  48  miles  an  hour.  Find  the 
rate  of  the  wind  and  the  rate  at  which  the  bird  can  fly  in 
calm  weather. 

24.  There  is  a  number  whose  two  digits  differ  by  2.  If  the 
digit  in  units'  place  is  multiplied  by  3  and  the  digit  in  tens' 
place  is  multiplied  by  2,  the  number  is  increased  by  44.  Find 
the  number,  the  tens'  digit  being  the  larger. 

25.  In  a  number  consisting  of  two  digits  one  digit  is  equal 
to  twice  their  difference.  If  the  order  of  the  digits  is  reversed, 
the  number  is  increased  by  18.  Find  the  number,  the  units' 
digit  being  the  larger. 

26.  If  the  length  of  a  rectangle  is  doubled  and  8  inches  added 
to  the  width,  the  area  of  the  resulting  rectangle  is  180  square 
inches  greater  than  twice  the  original  area.  If  the  length  and 
width  of  the  rectangle  differ  by  10,  what  are  its  dimensions  ? 

27.  The  Centigrade  reading  at  the  boiling  point  of  alcohol 
is  96°  lower  than  the  Fahrenheit  reading.  Find  both  the 
Centigrade  and  the  Fahrenheit  reading  at  this  temperature. 

Use  C  and  F  as  the  unknowns.  Then  one  of  the  equations  is  the 
formula  connecting  Fahrenheit  and  Centigrade  readings  obtained  on 
page  129,  and  the  other  is  C  +  96  =  F. 

28.  The  Centigrade  reading  at  the  boiling  point  of  mercury 
is  312°  lower  than  the  Fahrenheit  reading.  Find  both  the 
Fahrenheit  and  the  Centigrade  reading  at  this  temperature. 


164      INTRODUCTION   TO   SIMULTANEOUS  EQUATIONS 

29.  There  is  a  number  consisting  of  three  digits,  those  in 
tens'  and  units'  places  being  the  same.  The  digit  in  hundreds' 
place  is  4  times  that  in  units'  place.  If  the  order  of  the  digits 
is  reversed,  the  number  is  decreased  by  594.  What  is  the 
number  ? 

30.  A  man  rowing  against  a  tidal  current  drifts  back  2\ 
miles  per  hour.  Rowing  with  this  current,  he  can  make  121 
miles  per  hour.  How  fast  does  he  row  in  still  water  and  how 
swift  is  the  current  ? 

31.  Flying  against  a  wind  a  bird  makes  28  miles  per  hour, 
and  flying  with  a  wind  whose  velocity  is  2|  times  as  great,  the 
bird  makes  46  miles  per  hour.  What  is  the  velocity  of  the 
wind  and  at  what  rate  does  the  bird  fly  in  calm  weather  ? 

32.  A  freight  train  leaves  Chicago  for  St.  Paul  at  11  a.m. 
At  3  and  5  p.m.  respectively  of  the  same  day  two  passenger 
trains  leave  Chicago  over  the  same  road.  The  first  overtakes 
the  freight  at  7  p.m.  the  same  day,  and  the  other,  which  runs 
10  miles  per  hour  slower,  at  3  a.m.  the  next  day.  What  is 
the  speed  of  each? 

33.  Two  boys,  A  and  B,  having  a  30-lb.  weight  and  a  teeter 
board,  proceed  to  determine  their  respective  weights  as  fol- 
lows :  They  find  that  they  balance  when  B  is  6  feet  and  A  5 
feet  from  the  fulcrum.  If  B  places  the  30-lb.  weight  on  the 
board  beside  him,  they  balance  when  B  is  4  and  A  5  feet  from 
the  fulcrum.     How  heavy  is  each  boy  ? 

34.  C  is  6£  feet  from  the  point  of  support  and  balances  D, 
who  is  at  an  unknown  distance  from  this  point.  O  places  a 
33-lb.  weight  beside  himself  on  the  board  and  when  4|  feet  from 
the  fulcrum,  balances  D,  who  remains  at  the  same  point  as  be- 
fore. C's  weight  is  84  pounds.  What  is  D's  weight  and  how 
far  is  he  from  the  fulcrum  ? 

35.  E  weighs  95  pounds  and  F  110  pounds.  They  balance 
at  certain  unknown  distances  from  the  fulcrum.     E  then  takes 


PROBLEMS  IN   TWO    VARIABLES  165 

a  30-lb.  weight  on  the  board,  which  compels  F  to  move  3  feet 
farther  from  the  fulcrum.  How  far  from  the  fulcrum  was  each 
of  the  boys  at  first  ? 

36.  A  fast  freight  leaves  New  York  for  Chicago  at  8  a.m. 
At  4  p.m.  the  same  day  an  express  train  leaves  New  York  for 
Chicago  and  passes  the  freight  12  hours  later.  Another  ex- 
press leaving  New  York  at  6  p.m.  of  the  same  day  overtakes 
the  freight  10  hours  after  starting.  Find  the  rate  of  each 
train  if  the  second  express  goes  8  miles  per  hour  faster  than 
the  first. 

37.  The  Centigrade  reading  at  the  melting  point  of  silver 
is  796°  lower  than  the  Fahrenheit  reading.  Find  both  Centi- 
grade and  Fahrenheit  readings  at  this  temperature. 

38.  The  Fahrenheit  reading  at  the  melting  point  of  gold  is 
992°  higher  than  the  Centigrade  reading.  Find  both  Centi- 
grade and  Fahrenheit  readings  at  this  temperature. 

39.  $10,000  and  $8000  are  invested  at  different  rates  of 
interest,  yielding  together  an  annual  income  of  $  820.  If  the 
first  investment  were  $  12,000  and  the  second  $  6000,  the 
yearly  income  would  be  $  840.     Find  the  rates  of  interest. 

40.  In  a  switch  yard  a  car  weighing  50  tons  and  going  at  a 
certain  rate  strikes  a  standing  car,  whereupon  both  cars  move 
off  at  the  rate  of  4  miles  per  hour.  If  the  second  car,  moving 
at  the  same  rate  as  the  first  before  impact,  were  to  strike  the 
first  car  when  standing  still,  they  would  move  off  at  the  rate 
of  2  miles  per  hour.  How  fast  did  the  first  car  move  before 
impact  and  what  is  the  weight  of  the  second  car  ?  (Set  up  the 
equations  by  means  of  the  formula  iv^x  =  w-px  +  w2v2,  ob- 
tained on  page  127.) 

41.  200  ccm.  of  white  oak  is  fastened  to  25  ccm.  of  steel, 
making  a  combination  whose  average  density  is  1.56.  If 
250  ccm.  of  oak  is  fastened  to  20  ccm.  of  steel,  the  average 
density  of  the  combination  is  1.3.  Find  the  density  of  white 
oak  and  also  of  steel. 


166      INTRODUCTION  TO   SIMULTANEOUS  EQUATIONS 

SIMULTANEOUS   EQUATIONS   IN   THREE   VARIABLES 

120.  Illustrative  Problem.  Three  men  were  discussing  their 
ages  and  found  that  the  sum  of  their  ages  was  90  years.  If 
the  age  of  the  first  were  doubled  and  that  of  the  second  trebled, 
the  aggregate  of  the  three  ages  would  then  be  170.  If  the  ages 
of  the  second  and  third  were  each  doubled,  the  sum  of  the  three 
would  be  160.     Find  the  age  of  each  ? 

Solution.  Let  x,  y,  and  z  represent  the  number  of  years  in  their 
ages  in  the  order  named. 

Then,                                    x  +  y  +  z  =  90,  (1) 

2x+S y+z=  170,  (2) 

and                                     x+2y+2z=  160.  (3) 

Since  by  supposition  x  represents  the  same  number  in  all  three 
equations,  and  likewise  y  and  z,  if  we  subtract  (1)  from  (2),  we  obtain 
a  new  equation  from  which  z  is  eliminated. 

I.e.                                         x  +  2y=80.  (4) 

Again,  multiplying  (2)  by  2  and  subtracting  (3), 

3  x  +  4  y  =  180.  (5) 

(4)  and  (5)  are  two  equations  in  the  two  variables  x  and  y.  Solving 

these  by  eliminating  y,  we  find       x  =  20.  (6) 

Substituting  x  =  20  in  (4),          y  =  30.  (7) 

Substituting  x  and  y  in  (1),         z  =  40.  (8) 

Check  by  substituting  the  values  of  x,  y,  and  z  in  all  three  given 
equations  and  also  by  showing  that  they  satisfy  the  conditions  of  the 
problem. 

The  values  of  x,  y,  and  z  as  thus  found  constitute  the  solution 
of  the  given  system  of  equations. 

Evidently  x  could  have  been  eliminated  first,  using  (1),  (2)  and 
(1),  (3),  giving  a  new  set  of  two  equations  in  y  and  z.  Let  the 
student  find  the  solution  in  this  manner. 

Also  find  the  solution  by  first  eliminating  y,  using  (1),  (2)  and  (2), 
(3),  getting  two  equations  in  x  and  z,  from  which  the  values  of  x  and 
z  can  be  found. 


EQUATIONS  IN   THREE   VARIABLES 


167 


121.  Definition.  An  equation  is  said  to  be  of  the  first  degree 
in  three  variables  if  no  one  of  the  variables  is  multiplied  by 
itself  or  by  one  of  the  others  (§  110). 

The  fact  that  the  solutions  are  found  to  be  the  same  no 
matter  in  what  order  the  equations  are  combined,  indicates 
that  a  system  of  three  independent  and  simidtaneous  eqtiations  of 
the  first  degree  in  three  variables  has  one  and  only  one  solution. 

As  in  the  case  of  two  equations,  each  should  be  first  reduced 
to  a  standard  form  in  which  all  the  terms  containing  a  given 
variable  are  collected  and  united  and  all  fractions  removed  by  M, 
Principle  VIII. 

EXERCISES 

Solve  the  following  systems  of  equations,  and  check  the 
results  by  substituting  the  values  found  for  each  variable  in 
the  given  equations : 


2  x  —  y  +  z  =  18, 
x-2y  +  3z  =  10, 

3  x  +  y  -  4  z  =  20. 

5x  —  3y  -\-z  =  15, 
ix  +  3y—  z  —  3, 
2x  —  y  +  z  =  8. 

4:x  +  2y  +  z  =  13, 
x—y  +  z  =  4:, 
x  +  2y  —  z  —  \. 

6»+4y  —  4z  =  —  4, 
4a-2y  +  8z  =  0, 
x  +  y+z  =  4:. 

(x  +  2y  +  3z  =  5, 
5.    <4:X  —  3y  —  z  =  5, 

[x  +  y  +  z  =  2. 


9. 


10. 


\2x-8y-£3z  =  2, 
|X-4y  +  5z  =  l, 
[3x-10y-z  =  5. 

x  +  y  +  z  =  l, 
x  +  3y  +  2z  =  8, 
2x  +  8y-3z=:15. 

\2x  —  3y+z  =  5, 
3a  +  2y-z  =  5, 
{x  +  y  +  z  =  3. 

\x+  y  +  z  ==  6, 
3  x  _  2  y  -  z  =  13, 
[2x-y  +  3z  =  26. 

lx  +  y  +  z  =  6, 
Ax  —  y  —  z=  —  1, 
[2x  +  y-3z=-G. 


1G8      INTRODUCTION  TO   SIMULTANEOUS  EQUATIONS 

PROBLEMS   INVOLVING    THREE    VARIABLES 

122.  Illustrative  Problem.  A  broker  invested  a  total  of 
$15,000  in  the  street  railway  bonds  of  three  cities,  the  first 
investment  yielding  3  %,  the  second  3|  %,  and  the  third  4  %, 
thus  securing  an  income  of  $535  per  year.  If  the  second 
investment  was  one-half  the  sum  of  the  other  two,  what  was 
the  amount  of  each  ? 

Solution.  Suppose  x  dollars  were  invested  at  3  %,  y  dollars  at  3 \  %, 
and  z  dollars  at  4  %. 


Then,                                              x  +  y  +  z  =  15000, 

(1) 

.03  x  +  .035  y  +  .04  z  =  535, 

(2) 

and                                                              x  +  z  =  2  y. 

(3) 

From  (3),                                   ar-2^  +  z  =  0. 

(4) 

Subtract  (4)  from  (1),                          3  y  =  15000, 

(5) 

and                                                                  y  =  5000. 

(6) 

From  (1),   by  M,  .035  x  +  .035  y  +  .035  z  =  525. 

(7) 

Subtract  (7)  from  (2),     -  .005  x  +  .005  z  =  10. 

(8) 

Divide  (8)  by  .005,                            -x  +  z  =  2000. 

(9) 

Substitute  (6)  in  (4),                         x  +  z  =  10000. 

(10) 

Add  (9)  and  (10),                                  2  z  =  12000. 

(11) 

z  =  6000. 

(12) 

Substitute  (5)  and  (12)  in  (1),                 x  =  4000. 

(13) 

Hence,  $4000,  $5000,  and  $6000  were  the  sums  invested. 

Solve  the  following  problems,  using  three  unknowns : 

1.  The  sum  of  three  angles  A,  B,  and  O  of  a  triangle  is  180 
degrees.  |  of  A+\  of  B+\  of  C  is  48  degrees,  while  I  of  A+ \ 
of  B-\-\  of  C  is  30  degrees.    How  many  degrees  in  each  angle  ? 

2.  The  combined  weight  of  1  cubic  foot  each  of  compact 
limestone,  granite,  and  marble  is  535  pounds.  1  cubic  foot  of 
limestone,  2  of  granite,  and  3  of  marble  weigh  together  1041 
pounds,  while  1  cubic  foot  of  limestone  and  1  of  granite  together 
weigh  195  pounds  more  than  1  cubic  foot  of  marble.  Find  the 
weight  per  cubic  foot  of  each  kind  of  stone. 


PROBLEMS  IN   THREE   VARIABLES  169 

3.  A  number  is  composed  of  3  digits  whose  sum  is  7.  If 
the  digits  in  tens'  and  hundreds'  places  are  interchanged, 
the  number  is  increased  by  180 ;  and  if  the  order  of  the  digits 
is  reversed,  the  number  is  decreased  by  99.  What  is  the 
number  ? 

4.  The  sum  of  the  angles  A,  B,  and  C  of  a  triangle  is  180 
degrees.  If  B  is  subtracted  from  \  of  A  the  remainder  is 
\  of  C,  and  when  C  is  subtracted  from  twice  A  the  remainder 
is  4  times  B.     How  many  degrees  in  each  angle  ? 

5.  The  sum  of  the  three  sides  a,  b,  c  of  a  certain  triangle  is 
35,  and  twice  a  is  5  less  than  the  sum  of  b  and  c,  and  twice  c 
is  4  more  than  the  sum  of  a  and  b.  What  is  the  length  of 
each  side  ? 

6.  The  combined  number  of  students  at  Harvard,  Yale, 
and  Columbia  during  the  year  1905-1906  was  13,390.  The 
number  at  Harvard  minus  the  number  at  Columbia  plus  twice 
the  number  at  Yale  was  6893,  and  the  number  at  Columbia 
plus  4  times  the  number  at  Yale  minus  twice  the  number 
at  Harvard  was  7258.  What  was  the  number  at  each 
university  ? 

7.  The  total  number  of  students  at  the  universities  of 
Illinois,  Michigan,  and  Wisconsin  during  the  year  1905-1906 
was  12,216.  Twice  the  number  at  Illinois  plus  3  times  that  at 
Michigan  plus  4  times  that  at  Wisconsin  was  36,145.  If  the 
number  at  Michigan  is  subtracted  from  the  sum  of  the  num- 
bers at  Illinois  and  Wisconsin,  the  remainder  is  3074.  Find 
the  number  at  each  university. 

8.  The  total  number  attending  schools  in  the  United  States 
in  1904-1905  was  17,953,844.  If  the  number  in  secondary 
schools  and  colleges  combined  be  subtracted  from  the  number 
in  elementary  schools,  the  remainder  is  15,924,656;  while  if 
twice  the  number  in  colleges  be  added  to  the  number  in  ele- 
mentary and  secondary  schools  combined,  the  sum  is  18,092,388. 
Find  the  number  of  students  of  each  kind. 


170     INTRODUCTION  TO  SIMULTANEOUS  EQUATIONS 


9.  The  combined  foreign  trade  in  1905  at  the  three  ports, 
London,  Liverpool,  and  New  York,  was  3598  million  dollars. 
If  the  amount  at  London  is  subtracted  from  the  combined 
amounts  at  New  York  and  Liverpool,  the  remainder  is  988 
million ;  and  if  the  amount  at  New  York  is  subtracted  from 
the  combined  amounts  at  London  and  Liverpool,  the  remainder 
is  1384  million.  Find  the  amount  of  foreign  trade  at  each 
port. 

In  the  following  three  examples  find  the  number  of  seconds 
in  each  record  and  reduce  the  results  to  minutes  and  seconds. 

10.  If  x  is  the  number  of  seconds  in  the  Eastern  inter- 
collegiate record  for  a  mile  run,  y  the  number  in  the  Western 
intercollegiate  record,  and  z  the  number  in  the  world's  record, 
then 

x  +  y  +  z  =  781.15, 

—x+2y+z=  519.35, 

2  x  —  y  + 1  =  514.55. 

11.  If  a;  is  the  number  of  seconds  in  the  Eastern  inter- 
collegiate record  for  a  half  mile  run,  y  the  number  in  the  West- 
ern intercollegiate  record,  and  z  the  number  in  the  world's 
record,  then 

2x  +  3y  +  z  =  697.7, 

3x  +  2y  +  2z  =  809.8, 
2x-y  +  z  =  22S.l. 

12.  If  x  —  number  of  seconds  in  the  world's  mile  trotting 
record  in  1806,  y  =  number  of  seconds  in  the  world's  record  in 
1885,  and  z  =  number  of  seconds  in  the.  world's  record  in  1903, 
then 

*  +  y  +  z  =  426.25, 
2z  +  4?/  +  6z  =  1584, 
-  a>  +  y +  2  «  =  186.75. 


REVIEW  QUESTIONS  171 

REVIEW  QUESTIONS 

1.  How  may  a  point  in  a  plane  be  located  by  reference  to 
two  fixed  lines  ?  What  are  these  fixed  lines  called  ?  What 
names  are  given  to  the  distances  from  the  point  to  the  fixed 
lines  ?  Why  are  negative  numbers  needed  in  order  to  locate 
all  points  in  this  manner  ? 

2.  Draw  a  pair  of  axes  in  a  plane  and  locate  the  following 
points:  (5,0),  (-2,0),  (0,3),  (0,-1),  (0,0). 

3.  State  a  problem  involving  motion  and  solve  it  by  means 
of  a  graph. 

4.  How  many  pairs  of  numbers  can  be  found  which  satisfy 
the  equation  x  —  2  y  =  6  ?  State  five  such  pairs  and  plot  the 
corresponding  points.  How  are  these  points  situated  with 
respect  to  each  other  ?  What  can  you  say  of  all  points  corre- 
sponding to  pairs  of  numbers  which  satisfy  this  equation? 
What  is  meant  by  the  graph  of  an  equation  ? 

5.  How  many  pairs  of  numbers  will  simultaneously  satisfy 
the  two  equations  3x+2y=7  and  x +y  =  3 ?  Show  by  means 
of  a  graph  that  your  answer  is  correct. 

6.  Describe  elimination  by  the  process  of  substitution  ;  also 
by  the  process  addition  or  subtraction.  Under  what  conditions 
is  one  or  the  other  of  these  methods  preferable  ? 

7.  Why  is  the  solution  by  elimination  in  some  cases  pref- 
erable to  the  solution  by  means  of  the  graph  ? 

8.  Describe  the  solution  of  a  system  of  three  linear 
equations  in  three  unknowns.  Is  it  immaterial  which  of  the 
three  variables  is  eliminated  first  ? 

9.  Can  you  find  a  definite  solution  for  two  equations  each 
containing  three  unknowns  ?  Illustrate  this  by  means  of  the 
equations  4  a;  —  3y  —  z  =  5  and  x  +  y  +  z  =  2. 


CHAPTER  VI 
SPECIAL  PRODUCTS  AND   FACTORS 

123.  Repeated  Factors.  Number  expressions  containing  re- 
peated factors  have  already  been  considered  in  Chapter  III. 
x-x  was  written  x2  and  called  the  square  of  x,  or  x  square; 
similarly,  (a  +  b)  (a  -f  b)  was  written  (a  +  6)2  and  read  the 
square  of  the  binomial  a  +  6  or  the  binomial  a  +  b  squared. 

124.  Definitions.  Any  number  written  over  and  to  the  right 
of  a  number  expression  is  called  an  index  or  exponent  and,  if 
a  positive  integer,  shows  how  many  times  that  expression  is  to 
be  taken  as  a  factor. 

A  product  consisting  entirely  of  equal  factors  is  called  a 
power  of  the  repeated  factor.  The  repeated  factor  is  called  the 
base  of  the  power. 

E.g.  x%  means  x  •  x  -x  and  is  read  the  third  power  of  x  or  x  cube;  x6 
means  x  •  x  •  x  •  x  •  x,  and  is  read  the  fifth  power  of  x  or  briefly  x  fifth, 
(x  —  y)s  =  (x  —  y)(x  —  y)(x  —  y)  and  is  read  (x  —y)  cubed  or  the 
cube  of  the  binomial  (x  —  y). 

Notice  two  important  differences  between  an  exponent  and 
a  coefficient. 

(1)  5a  =  a+a-r-a  +  a  +  a,  while  a5  =  a  •  a  •  a  •  a  •  a. 

(2)  In  5  abc  the  coefficient  5  applies  to  the  product  abc, 
while  in  abc5,  the  exponent  5  applies  only  to  the  factor  c.  In 
order  to  make  it  apply  to  the  product  it  is  necessary  to  use  a 
parenthesis,  thus,  (abc)5  means  the  product  abc  taken  five 
times  as  a  factor. 

172 


PRODUCTS  OF  POWERS  OF  SAME  BASE 


173 


EXERCISES 

Perform  the  following  indicated  multiplications 


1. 

23 

,  24,  25, 

26. 

9. 

102,  103, 104, 105. 

17. 

(x  —  y  —  5)2. 

2. 

32 

33,  3*, 

35. 

10. 

(a  +  bf. 

18. 

(w  +  z  —  4)2. 

3. 

42 

43,  4«, 

45. 

11. 

(c-d)\ 

19. 

(2—z  —  x  +  w)2. 

4. 

52 

53,  54, 

55. 

12. 

982  =  (100-2)2. 

20. 

(x  —  y  —  5)2. 

5. 

62 

63,  64. 

13. 

(a  +  b  +  cy. 

21. 

(a  -  y)\ 

6. 

72 

7s,  74. 

14. 

(a  +  6  -  c)2. 

22. 

(x  +  y)3. 

7. 

82 

83,  8*. 

15. 

(3  -  a)2. 

23. 

(a  +  b)\ 

8. 

92 

9s,  94. 

16. 

(3  _  6  _  C)2. 

24. 

(c  -  d)4. 

125 

.   In  the  case 

of  factors  expressed 

in  Arabic  figures  mul- 

tiplications  like  the  following  may  be  carried  out  in  either  of 
two  ways. 

E.g.  32  •  3*  =  9  •  81  =  729. 

or  32  •  34  =  (  3  •  3)(3  •  3  •  3  •  3)  =  32+4  =  36  =  729. 

But  with  literal  factors  the  second  process  only  is  possible. 
E.g.  a2. a*  =  (a  •  d)(a  •  a  •  a  •  a)  =  a?+*  =  a6. 

The  process  in  which  the  exponents  are  added  applies  only 
when  the  factors  are  powers  of  the  same  base. 

E.g.  28  .  32  =  8  •  9  =  72  cannot  be  found  by  adding  the  exponents. 

21  or  2  is  called  the  first  power  of  2. 
Thus  2  •  28  =  21  •  28  =  21+8  =  24. 


EXERCISES 

In  the  following  exercises  carry  out  each  indicated  multipli- 
cation in  two  ways  in  case  this  is  possible : 

1.  25.26.  3.   2-24.  5.   3-34.  7.   42-43. 

2.  22-23.  4.    32-33.  6.   32-35.  8.   4-44. 


174 


SPECIAL   PRODUCTS  AND  FACTORS 


9.  5-52. 

12.   7-73. 

15.    x7-x\ 

18.    23-22.24. 

10.   52-58. 

13.    a2 -a3. 

16.    *3-t4. 

19.   3-32-33. 

11.   62-62. 

14.   tf-x2. 

17.    J2-*3-*4. 

20.   22.23-22.2 

126.  Illustrative  Problem.  To  multiply  2*  by  2n,  k  and  n 
being  any  two  positive  integers. 

Solution.    2*  means  2  •  2  . 2  •  2,  etc.,  to  k  factors, 
and  2n  means  2  •  2  •  2  •  2,  etc.,  to  n  factors. 

Hence  2* .  2»    =  (2  . 2  .  2  • . .  to  n  factors)  (2  •  2  •  •  •  to  i  factors) 

=  2.2.2.2...  to  fc  +  n  factors  in  all. 
That  is,  2*.2»=2*+». 

The  preceding  examples  illustrate  the  following  principle : 

127.  Principle  XIV.  The  product  of  two  powers  of  the 
same  base  is  found  by  adding  the  exponents  of  the  powers 
and  making  this  sum  the  exponent  of  the  common  base. 


EXERCISES 

Perform  the 

following 

indicated 

multiplications  by  means 

of  Principle  XIV : 

1.   23-27. 

8. 

32« .  3» 

15.   wx-wy+Sx. 

2.   a3  •  a7. 

9. 

52+»  ,  52-» 

16.   n2-n3c+ih. 

3.   34-35. 

10. 

J2x+v   .  J2x 

~".              17.   cx  •  c2_x. 

4.    xA  •  Xs. 

11. 

am  •  an.  . 

18.   ar-z40. 

5.   3*. 3". 

12. 

/2a  .  fb+a 

19.    r30-?'40. 

6.    sc*.zB. 

13. 

y*a  -  y3"- 

20.    s^-s™. 

7.   4a  •  4J 

14. 

qAb  ,  rtfa+ib 

21.    -y2*"1  .  <y3*+2# 

Perform  the  following  multiplications  by  means  of  Princi- 
ples IV,  III,  and  XIV : 

22.  23(22  +  24).  25.    aX^b-ab3). 

23.  22(3  •  24  +  5  •  2s).  26.   a^a(4  of  +  3  a88). 

24.  44(3-43-5.42).  27.   ^(5^-3^). 


PRODUCTS   OF  MONOMIALS  175 

Multiply  the  following  and  state  all  principles  used : 

28.  a2m2  -  b2m3  by  m4.  37.  32  ■  42n  -  63  •  44n  by  43\ 

29.  4.32-5-7-2by  24.  38.  aV-ac^byafc. 

30.  2-3-4-32-i-7.3by  3.  39.  62  •  43-74  •  42  +  44  by  43. 

31.  4z2-3ar3  +  6a4by  x2.  40.  12 x2f  -  6 x?y2  +  3  xy  by  x3. 

32.  ox-3^  +  2»4by  x4.  41.  2  a36 - 3 a2c - 4 a45  by  a2*. 

33.  3  y2  +  4  y*  -  y3  by  y.  42.  6  ar3"-25  +  8  x2"-3*  by  x2a+2i. 

34.  7x4-5x?-2xby  &.  43.  y3m+2n - y2n  3m+l  by  ^m+3n. 

35.  3  a2&4  +  2  d2b  -  4  a262  by  a3.  44.  ar5c-3a  +  ark+2a  +  a;2abyz4a-2c. 

36.  4  a26  —  a;?c  +  a2d  by  a26.  45.  a"*-**  —  abx-,lx  —ax  by  a5*+n*. 

PRODUCTS   OF   MONOMIALS 
128.    In  multiplying  two  numbers  each  of  which  is  in  the 
factored  form,  if  the  factors  are  all  expressed  in  Arabic  figures, 
the  operation  may  be  carried  out  in  two  ways. 

E.g.  (2  •  3)  (2  •  3  •  5)  =  6  •  30  =  180. 

Also  (2  •  3)  (2  •  3  •  5)  =  22  •  32  •  5  =  4  •  9  •  5  =  180. 

In  the  second  process  one  of  the  factors,  2  or  3,  is  multiplied  in 

and  this  product  is  then  multiplied  by  the  other  factor,  2  being 

combined  with  the  factor  2  and  3  with  3  by  Principles  III  and  XIV. 

In  the  case  of  literal  factors  the  second  process  only  is  available. 

E.g.  (a262)  (5  a*b»c)  =  5  a4+2  62+»  c  =  5a%5c. 

EXERCISES 

Perform  the  following  indicated  multiplications,  each  in  two 
ways  where  possible : 

1.  (5  •  72)(53  •  7s).  4.    (3.42)(4-32).        7.   6t(S$P). 

2.  (32-5)(3.23.52).       5.   3a6(5a262).  8.   7m(4mV). 

3.  (4  •  53)(5  •  43).  6.   4  x(3  xy).  9.    6  k2(62tf). 

By  definition  (§  77),  expressions  in  the  factored  form,  such 
as  3  ab,  5  a2b2,  3  •  23  •  52,  etc.,  are  monomials. 


176  SPECIAL  PRODUCTS  AND  FACTORS 

The  preceding  examples  illustrate  the  following  principle : 

129.  Principle  XV.  The  product  of  two  monomials  is 
found  by  multiplying  either  one  by  each  factor  of  the 
other  in  succession. 

Each  factor  of  the  multiplier  is  associated  with  any  desired  factor 
of  the  multiplicand  according  to  Principle  III,  and  where  the  bases 
are  the  same  the  exponents  are  added  according  to  Principle  XIV. 

If  there  are  no  factors  common  to  multiplier  and  multiplicand,  the 
product  can  only  be  indicated  by  writing  all  the  factors  of  both  in 
succession. 

Multiply :  exercises 

1.  4- 7 -8-9  by  2 -3 -5.  5.  3  *V  by  4  ay*. 

2.  3  .8-  2  by  2-  5-  6.  6.   5  a2b3c  by  ab*c. 

3.  2  xyz  by  3  z?yz.  7.    2  a^&V  by  5  xb*c. 

4.  63  •  2*  •  x5  by  6  •  23 .  x.  8.    9  x° yV  by  4  x2ay2hzc. 

9.    6  x*n+lfn~hn  by  3  xl-*ny5~n^~n- 

10.  3  n2x+4my-5r21-1  by  w^m'-V*. 

In  each  of  the  following  exercises  state  which  of  the  Prin- 
ciples I-XV  are  used : 

11.  32-23(24.32-4.22-2.32). 

12.  6.34(42-22.32+43.24-5.34). 

13.  2x(4:  +  7xi-3x2y),  16.   4*y(3 <m— 4  bg +2 xy). 

14.  4yz(3y2x3  —  fx2 -{- y*x*) .        17.   2xa(xa  —  xb  —  3c). 

15.  5  a2b\as  -  b3  +  a2b2).  18.    2  yaf  (y  -  as*  +  3  y). 

19.  4a^,+X^n+1-4y2M+5  +  5a2ny4). 

20.  4  a?!n+y,+2m(y,2/m  —  ccy  +  £"*  +  yn). 

21.  (a  +  by(a-b).  24.    (a3+a26  +  a62+63)(a-6). 

22.  (a  -  6)3(a  +  6).  25.    (a  +  *  —  c)  (a+  &  +  c). 

23.  (a  +  6)2(a-&)\  26.    (3a>-2y-l)(2a>  +  y). 


MONOMIAL  FACTORS 


177 


27. 

(5a-3&)(6a2  +  262-l). 

39.    (i 

28. 

(3a2-262-3)(4a  +  3  63). 

40.    (i 

29. 

(1  +  a  +  «2)  (1  -  a). 

41.    (: 

30. 

(1  -  a  +  a2  -  a3)  (1  +  a). 

42.    ( 

31. 

(a  +  &)(a-6)(a2  +  62). 

43.    ( 

32. 

(a  +  6)  (a2 -aft  +  6s). 

44.    (. 

33. 

(a  -  b)  (a2  +  a&  +  b2). 

45.    (i 

34. 

(x-\-y)(x  —  y). 

46.    (i 

35. 

(100  + 1)  (100  - 1). 

47.    0 

36. 

(100  +  2)  (100  -  2). 

48.    (i 

37. 

(x  +  3)  (cc  +  5). 

49.    (i 

38. 

(a?  -  3)  (a?  -  5). 

50.    (< 

(t  +  $)(t-3). 
(x-2y)(x  +  3y). 
(x*  +  x  +  l)(x*-x  +  l). 
(a;  +  y)  (a3—  a;2?/  +  xy2—f). 
(x  -y)  (tf+xty+xf+f). 
(x2  -  xy  +  y2)  (x2  +  an/  +  /). 
(ar'  +  2  any  +  */2)  (a;  -  y)2. 
(tf-y^^  +  xttf  +  y*). 
(x2  +  y^(x*-x2y2  +  y*). 
(x2  +  y2)  (x  +  y)  (x  -  y). 
(x—y)  (xf+xy+y2)  (ar'+y5)- 
(an-  6")  (a2rt-f  an6n.+  Z>2"). 


FACTORS  OF  NUMBER  EXPRESSIONS 

130.  The  factors  of  numbers  are  of  great  importance  in 
arithmetic.  For  instance,  the  multiplication  table  consists  of 
pairs  of  factors  whose  products  are  committed  to  memory  for 
constant  use.  Likewise  in  algebra  the  factors  of  certain  special 
forms  of  number  expressions  are  so  important  that  they  must 
be  known  at  sight. 

An  expression  containing  no  fractions  is  said  to  be  prime  if 
it  has  no  integral  factors  except  itself  and  1. 

Thus  2,  3,  x,  x  +  2,  a2  +  b2,  are  prime  expressions. 

A  prime  expression  may  be  factored  by  using  fractions  or 
radicals.     See  p.  214. 


Thus 


5  =  2-2^  and  2=  V2-  V2- 


Such  factors  are  not  included  in  what  are  here  called  prime 
factors. 


178  SPECIAL  PRODUCTS  AND  FACTORS 

131.  Monomial  Factors.  If  the  terms  of  a  polynomial  con- 
tain a  common  factor,  they  may  be  combined  with  respect 
to  this  factor  according  to  Principles  I  and  II.  The  ex- 
pression is  thus  changed  into  a  product  of  a  monomial  and 
a  polynomial. 

Illustrative  Examples. 

1.  ax  +  ay  =  a(x  + y),  by  Principle  I. 

2.  a2  —  ab  =  a(a  —  b),  by  Principle  II. 

3.  9.8  +  3.4.5  =  3.4(3.2+5),  by  Principle  I. 

4.  6  a2b  -  4  ab2  =  2  ab  (3  a  -  2  b),  by  Principle  II. 

5.  5  xy  —  3  x2y  +  4  x*y  =  xy  (5  —  3  x  +  4  x2),  by  I  and  II. 

Observe  that  factoring  each  term  of  a  polynomial  does  not 
factor  the  polynomial. 

E.g.  330  -  210  -  60  is  not  factored  by  writing  it  2  •  3  •  5  •  11 
—  2> 3.5.7  + 22 .3-5;  but  by  adding  the  coefficients  of  the  common 
factor  2  •  3  •  5,  thus, 

2.3.5.11 -2- 3- 5- 7  + 22 .3.5  =  2. 3- 5  (11 -7  +  2). 

Likewise  10  a2bc  —  15  ab2c  +  20«6c2  is  not  factored,  although  each 
term  is  in  the  factored  form.  But  if  10  a?bc  —  15  ab2c  +  20  abc2  is 
written  in  the  form  5abc  (2  a  —  3  b  +  4  c),  it  is  then  factored. 

These  examples  illustrate  the  factoring  of  a  polynomial 
when  it  contains  a  monomial  factor  common  to  every 
term. 

When  such  a  common  factor  has  been  found  the  whole  ex- 
pression is  then  written  in  the  form  of  a  product  by  means  of 
Principles  I  and  II.  The  result  may  be  checked  by  Princi- 
ples IV  and  XV.  If  the  terms  of  a  polynomial  which  is  to  be 
factored  contain  a  common  factor,  this  should  always  be  re- 
moved at  the  outset. 


TRINOMIAL   SQUABES  179 

EXERCISES 

Factor  the  following  polynomials  : 

1.  8-12-18  +  48.  7.    15  xy*  -  20  afy  +  x>y\ 

2.  3-4- 5-15-20  +  35.  8.   9 vsw*  +  21  v4w3 - 18 v2w2. 

3.  3- 11- 4 -22- 2 +  44 -6.  9.   12  a*b3  -  8  cfb4  -  6  a2b2. 

4.  34-24-3-23-5.22.  10.   32.34-64.33-16-23.32. 

5.  6.54.7+3-53.2-54.9.22.  11.   27  •  23  +  54  •  24-36  •  23. 

6.  13  a*b  - 16  a3b*  -  2  a3b\  12.    11  aV  -  44a3x4  +  33ax. 

13.  1.2- 3-4  +  2- 3- 4- 5- 3- 4- 5- 6+4- 5- 6. 

14.  72  x4b3a  -  36  x*b2a2  -  48  xWa\ 

15.  84,  x9b7y4  + IS  x3bsy*  + 12  a^by. 

16.  19  •  34  •  24  •  53  -  13  •  34  •  23 .  52  -  29  ■  32  •  22 .  54. 

17.  17  a463c2  +  51  aW  -  34  aW. 

18.  38  a12  614c4  -  76  au&12c3  -  76  aM&uc». 

19.  4,x2aySb-^Qx3ay2b-8x'iayib. 

20.  3  a4n+263n+4  +  6  a6n+463n+3  - 12  a5n+365n+6. 

TRINOMIAL  SQUARES 
132.    In  §§  87  and  88  we  found  by  multiplication : 

{a  +  b)2  =  a2  +  2ab  +  b\  (1) 

and  (a  -  b)2  =  a2  -  2  ab  +  62.  (2) 

By  means  of  these  formulas  we  may  square  the  sum  or  dif- 
ference of  any  two  number  expressions. 

E.g.  (3x+2y)2=(3xy+2'3x.2y+(2y)2=9x2+12xy+4;y2. 
Also[(5  +  r)-(s-l)]2=(5  +  r)2-2(5  +  r)(s-l)  +  (s-l)2. 

The  last  expression  may  now  be  reduced  by  performing  the 
indicated  operations. 


180  SPECIAL  PRODUCTS  AND  FACTORS 

In  this  manner  write  the  squares  of  the  following  binomials. 
Verify  the  first  ten  results  by  actual  multiplication. 


1. 

t  +  8. 

11. 

7(a-3)-2(6  +  c). 

2. 

r-12. 

12. 

±(x-ij)  +  2(z  +  tf). 

3. 

Sx  —  5y. 

13. 

(3a-26)  +  5. 

4. 

6a-  7. 

14. 

7  x  —  (4  r  —  s). 

5. 

m5  +  3». 

15. 

7(m2-3)-6(m3  +  rc). 

6. 

Itf  +  Zx. 

16. 

3(z  +  y)-2(3  +  x). 

7. 

3tf-2y\ 

17. 

4(ar5-3y)  +  (ar!  +  4y2). 

8. 

8  m2n  —  7  win2. 

18. 

2x2y  —  b  (x  —  y). 

9. 

5a3-363. 

19. 

3(5x2-z)-4:x'z2. 

10. 

4  aWy3  —  3  a?xy*. 

20. 

7(r-s)  +  3(rV-2rs). 

133.  The  binomial  a  +  6  is  one  of  the  two  equal  factors  of 
a?  +  2ab  +  b2,  and  is  called  the  square  root  of  this  trinomial. 

Likewise  a  —  b  is  the  square  root  of  a2  —  2  ab  +  &2- 

In  each  case  a  is  the  square  root  of  a2  and  b  of  ft2.  Hence 
2  ab  is  twice  the  product  of  the  square  roots  of  a2  and  b2. 

From  the  squares  obtained  in  the  last  article,  we  learn  to 
distinguish  whether  any  given  trinomial  is  a  perfect  square, 
as  in  the  following  examples : 

1.  sc2  -}-  4  #  +  4  is  in  the  form  of  (1),  since  x2  and  4  are  squares 
each  with  the  sign  +,  and  4  a;  is  twice  the  product  of  the  square 
roots  of  x2  and  4.     Hence 

a?  +  4«  +  4  =  x2  -f  2  (2a?)  +  22  =  (a?  +  2)  (x  +  2)  =  (x  +  2)2. 

2.  cc2  —  4  a;  +  4  is  in  the  form  of  (2),  since  it  differs  from  (1) 
only  in  the  sign  of  the  middle  term.     Thus 

^-4^  +  4  =  ar!-2(2a0+22=  (as  -  2)  (x-2)=  (x  -  2)2. 

A  trinomial  which  is  the  product  of  two  equal  factors  is 
called  a  trinomial  square. 


TRINOMIAL   SQUARES  181 

EXERCISES 

Determine  whether  the  following  are  trinomial  squares,  and 
if  so  indicate  the  two  equal  factors.  If  any  trinomial  is  not  a 
square,  make  it  so  by  modifying  one  of  its  terms. 

1.  x2  +  2xy  +  y2.  10.    64  +  t2  -  16 t. 

2.  x2  -2xy-\-  y2.  11.    16  +  x2  —  8  x. 

3.  x4  +  2  arty2  +  y4.  12.9-6?/  +  y2. 

4.  a;4  -  2  arty2  +  y4.  13.    25ar!  +  16?/2  +  40ar?/. 

5.  m2  +  n2  —  2 mw.  14.    4m2+«2+2 m». 

6.  r2  +  s2  +  2rs.  15.    100  +  s2  +  20s. 

7.  4a^-8^  +  4?/2.  16.   64  +  49  +  112. 

8.  a6  +  &6  +  2  a363.  17.    16  a2  +  25  62  -  50  a&. 

9.  a8  +  68-2a464.  18.    16  a2  +  25  b2  +  40  a6. 

19.    81-  270&  +  225  62. 

134.  From  the  foregoing  examples  we  see  that  a  trinomial  is 
a  perfect  square  if  it  contains  two  terms  which  are  squares 
each  with  the  sign  +,  while  the  third  term,  whose  sign  is 
either  +  or  — ,  is  twice  the  product  of  the  square  roots  of  the 
other  two.  Then  the  square  root  of  the  trinomial  is  the  sum 
or  the  difference  of  these  square  roots  according  as  the  sign 
of  the  third  term  is  +  or  — . 

Since  on  multiplying  we  find  (a  —  6)2  and  (b  —  a)2  give  the  same 
result,  we  may  write  the  factors  of  a2  —  2  <xb  +  b2  either  (a  —  b)  (a  —  b) 
or  (b  —  a)(b  —  a).     See  page  214. 

EXERCISES 

Factor  the  following.  If  any  one  of  the  trinomials  is  found 
not  to  be  a  square,  make  it  so  by  modifying  one  of  its  terms. 

1.  9  +2-3-4  +  16.  3.   9x2 +  I8xy  +  9y2. 

2.  x2  +±y2  +4xy.  4.    4  x2  +  4  xy  +  y2. 


182  SPECIAL  PRODUCTS  AND  FACTORS 

5.  4  x2  +  8  xy  +  4  y2.  18.  81  a2-  216  a  +  144. 

6.  25x2  +  12xy  +  4:y2.  19.  4  a2  +  8  a62  +  462. 

7.  16ar9  +  16a*/  +  4?/2.  20.  9  64  +  18  62c4  +  9  c8. 

8.  9  r2  +  36  rs  +  25  s2.  21.  4  x2  +  4  y2  -  8  ay. 

9.  16x6  +  8x*y  +  2/2.  22.  9 a2  -  16a6  +  4 62. 

10.  4a8  +  12z4a3  +  9a4.  23.  9z4  -  24^6  +  1662. 

11.  a10  +  6 a56  +  9  62.  24.  25  +  49 z2  -  70a. 

12.  (a  +  l)2  +  2(a  +  1)6  +  62.  25.  -  30  a62  +  9  a2  +  25  64. 

13.  (z  +  3)2+4(;c  +  3)2/  +  4?/2.  26.  16  a2  -  24  a6  +  9  62. 

14.  ^  +  12 3* +  36.  27.  36  or*  -  84  a:  +  49. 

15.  a4 +  18  a2 +  12.  28.  25-90  +  81. 

16.  121  +  4  a8  -  44  x4.  29.  64  a?2  -  32  a;  +  9. 

17.  16x4  +  642/4-64^2.  30.  (3  +  a)2  +  62-26(3+a). 

31.  (2  -  a)2  -  2(2  -  x)  (x  -1)  +0  -  l)2. 

32.  (2  +yf  +  2(2  +  y)(l  +4)  +  (1  +  4)2. 

33.  (3a-26)2-10(3a-26)  +  25. 

34.  (6a-6)2+  (2a  +  l)2-2(6a-6)(2a  +  l). 

35.  25(a  +  6)2  +  50(a  +  6)  (a  -  6)  +  25(a  -  6)2. 

36.  x?  +  12  x(a  +  6  +  c)  +  36  (a  +  6  +  c)2. 

37.  49 (m  -  3)4  +  36 (m  +  I)6  -  84(m  -  3)2(m  +  l)3. 

38.  W(x  -  y)2  -  16  (a;  -  y)  (x  +  y)  +  4(a;  +y)2. 

39.  -  30(a  +  6)  (a  -  6)2  +25 (a  -  6)4  +  9(a  +  6)2. 


DIFFERENCE  OF  TWO   SQUARES 


183 


THE  DIFFERENCE  OF   TWO  SQUARES 
135.    By  multiplication, 

(a  +  b)  (a  -  b)  =  a2  -  b\ 

Translate  this  formula  into  words.  By  means  of  this  formula 
the  product  of  the  sum  and  difference  of  any  two  number 
expressions  may  be  found. 

E.g.     (3  a  +  2  b)  (3  a  -  2  b)  =  (3  a)2  -  (2  6)2=9  a2  -  4  A2. 
Also  (a;  +  y  -  z)  (x  +  y  +  z)  =  [(x  +  y)  -  z]  [(x  +  y)  +  z] 

The  last  expression  may  now  be  simplified  by  performing 
the  operations  indicated. 

In  this  manner  form  the  following  products.  Verify  the 
first  ten  by  actual  multiplication. 

1.  (4a  +  5ft)(4a-5  6).  9. 

2.  (24  x  +  12  y)  (24  x  -12 y).      10. 

3.  (16a2ft3-3c)(16a2ft3+3c).     11. 


xm+  yn)  (xm  —  yn). 

2n+l_i_£>2n-l\  (tfn+l ffn-\\ 


4.  (5 -6Z2)  (5  +  6*2).  12. 

5.  (3x-  2y)  (Sx  +  2y).  13. 

6.  (a8  -  f)  (a3  +  f).  14. 

7.  (Sa^-S^XSa^  +  Sy4).  15. 

8.  (>+(y-«)]0-(y-«)].  16. 


-(a-6)][c+(«-6)]. 

«-(#  +  «)][>+&  +  «)]. 
a  +  ft  +  c)(a  —  ft  —  c). 
a  —  6  +  c)  (a—  6  —  c). 
r-y-z)  (r-y  +  z). 
a  +  ft  +c)  (a  +  6  —  c). 


17.  [a  +  ft+(c-ef)][a+&-(c  —  d)]. 

18.  0  +  y  +  (w+v)][>+?/-(w  +  v)]. 

19.  [4  x  -  (a  -  2  ft)]  [4  x  +  (a  -  2  ft)]. 

20.  [a.+  2b-(x-y*)][a+2b+(x-y*)l 

21.  (11  b3x  -  3  bx>)  (11  b3x  +  3  ftz3). 


184  SPECIAL  PRODUCTS  AND  FACTORS 

From  the  preceding  examples  we  see  that  every  binomial 
which  is  the  difference  between  two  perfect  squares  is  com- 
posed of  two  binomial  factors ;  namely,  the  sum  and  the 
difference  of  the  square  roots  of  these  squares. 

E.g.  16  x2  —  9  y2  is  the  difference  between  two  squares,  (4  x)2  and 
(3  y)2.     Hence  we  have 

16x2-9#2  =  (4x)2-  (3y)2  =  (4x  +  3*/)(4x-3?/). 

EXERCISES 

Factor  each  of  the  following : 

1.  x'-Ay2.  8.  a2-l.  15.  4:-(x-2y)2. 

2.  9a?-36y*.  9.  l-9x4.  16.  16a-25a&2. 

3.  x*-b2.  10.  4 -36 a2.  17.  49 a -4 a?/2. 

4.  4ar>-9&8.  11.  1-64  a8.  18.  225-64afy4. 

5.  16a4-9Z>4.  12.  I44z2&4-1.  19.  576 a -144a?/2. 

6.  64 -b2.  13.  256 a*b6-c\  20.  58-38. 

7.  1-&2.  14.  l-(x  +  y)2.  21.  x*-8ly2. 

136.  It  is  important  to  determine  whether  a  given  expres- 
sion can  be  written  as  the  difference  of  two  squares. 

E.g.  a2  +  b2  +  2  ab  -  c2  =  (a  +  b)2  -  c2  =  (a  +  b  +  c)  (a  +  b  -  c). 
Also,  c2  -  a2  +  2  ab  -  b2  =  c2  -  (a2  -  2  aft  +  62)  =  c2  -  (a  -  6)* 

=  (c  —  a  +  b)(c  +  a  —  b). 

EXERCISES 

In  the  following  determine  whether  each  is  the  difference  of 
two  squares,  and  if  so,  factor  it  accordingly ;  if  not,  make  it 
so  by  modifying  one  of  its  terms. 

1.  ^-(y-z)2.  5.    4a2&2-(a2  +  62-c2)2. 

2.  (x-yf-z\  6.    a2-(62+c2  +  2  6c). 

3.  a2  +  &2_2a5_4.  7.    (2a-5)2-(3a  +  l)2. 

4.  a?  +  y2-2xy-z2.  8.    (3 x2 - yf  -  (x  +  y)2. 


THE  SUM  OF  TWO   CUBES  185 

9.  (3a-2  6)2-(8a  +  5&)2.  18.   9a2  +  6ab  +  b2 -c2. 

10.  (3m_4)2-(2m  +  3)2.  19.  16aj*-as+4a6  -462. 

11.  (2r  +  s)2-(3r-s)2.  20.  a2-2a&  +  62-c2. 

12.  81-(a  +  6+c)2.  21.  c2-(a2  +  2a6  +  62). 

13.  x*  +  ^  +  i_4a2.  22.  <?-(a2-2ab  +  b2). 

14.  a2-0  +  2i/)2.  23.   (a  +  6)2-(a-6)2. 

15.  9  s2 -(a -ft)8.  24.  a2  +  4a6  +  62-a^. 

16.  25m2-  (3  r  +  2s)2.  25.  a2  +  4a& +  452- x2. 

17.  4c2-(4a2  +  12a6-9  62).  26.  9a2  +  16  b2 -  32ab-x*. 

27.  a2  +  ±ab  +  4:b2  —  (x2  —  2xy  +  yi). 

28.  (3a;-2)2-(4^  +  9?/2-12a^). 

29.  a^  +  4a;y-i/2-(a2  +  2a6  +  62). 

30.  16  xAy2-  (4  a2 +  9/ +  12  ay). 

31.  (a  +  6)2-(4a2  +  9&2-12  6c). 

THE    SUM   OF   TWO  CUBES 
137.   By  multiplication  we  find 

(a  +  6)(a2-a6+62)  =  a3+63. 

The    pupil  should  perform   the   multiplication  indicated   in   this 
formula  and  carefully  note  how  the  terms  cancel  in  the  product. 

By  means  of  this  formula  obtain  the  following  products : 

1.  (x  +  y)(x2-xy  +  y2).  6.    (x2  +  y2)  (x*  -  x?y2  +  y*). 

2.  (a  +  2)(a2-2a  +  4).  7.    (5  a  +  b)  (25  a2  -  5  ab  +  b2). 

3.  (3a  +  6)(9a2-3a&+62).         8.   (1  +  oz2)(l  -5x2-\-25xi). 

4.  (l  +  4cc)(l-41r+16arJ).  9.    (1  +  2xy)(l-2xy+4:X2y2). 

5.  (w>2+3a2)04-3a2w;2+9a4).    10.    (2x+3y)(4;X2-6xy+9y2). 


186  SPECIAL  PRODUCTS  AND  FACTORS 

These  products  show  that  every  binomial  which  is  the  sum 
of  the  cubes  of  two  numbers  is  the  product  of  two  factors,  one 
of  which  is  the  sum  of  the  numbers,  and  the  other  is  the  sum 
of  their  squares  minus  their  product. 

E.g.  (1)  x*  +  y*  =  (x  +  y) (x2  -  xy  +  rf). 

(2)  8  a3  +  27  bs  =  (2  a)3  +  (3  b)s 

=  (2a  +  3J)(4a2-2a.36+  9  b2). 

(3)  X«  +  y*  =  (X2)8  +  (,,2)3  _.  (X2  +  ^  (x4  _  xy  +  ^4). 

.     Notice  the  difference  between  the  trinomial  x2  —  xy  +  y2  and  the 
trinomial  square  x2  —  2  xy  +  y2. 

EXERCISES 

Determine  whether  each  of  the  following  is  the  sum  of  two 
cubes,  and  if  so,  find  the  factors ;  if  not,  make  it  so  by  modi- 
fying one  of  its  terms.     Check  the  results  by  multiplication. 

1.  tf  +  tf.  9.  27a^  +  l.  17.  125  a?  +  y\ 

2.  a3  +  8  63.  10.  8a3  +  27  63.  18.  1  +  rr8. 

3.  27a3  +  63.  11.  8a3  +  64&3.  19.  64^  +  27^. 

4.  8a3  +  l.  12.  wV  +  ar>a3.  20.  83+103. 

5.  1  +  64  ar5.  13.  1  +  8  a?bs.  21.  1  +  729  a*. 

6.  23  +  33.  14.  64  a-5 +  343.  22.  »6  +  y12. 

7.  125  +  729.  15.  1  +  a3.  23.  a9  +  Z>3. 

8.  1  +  125  a6.  16.  a3 +  9  ft3.  24.  27^  + 125  s3. 

THE  DIFFERENCE   OF  TWO  CUBES 
138.  By  multiplication,  we  find 

(a  -  b)(a2  +  ab  +  b2)  =  a3  -  b5. 

In  performing  this  multiplication,  note  carefully  how  the 
terms  cancel  in  the  product. 


THE  DIFFERENCE  OF  TWO   CUBES  187 

By  means  of  this  formula  obtain  the  following  products. 

1.  (x-y^tf  +  xy  +  y-).  6.   (w2-2a2)(w4+2io2a2  +  4a4). 

2.  (&-3)<7>2  +  3Z>+9).  7.   (3a-2&)(9a2+6a&+4&2). 

3.  (2a-&)(4a2  +  2a&+62).   8.  (3a2-l)  (9x4  +  3a2  +  l). 

4.  (l-4«)(l+4aj  +  16aj2).     9.   (1  - 2 a#)  (1  +  2 ajy  +  4 ofy2). 

5.  (a2-62)(&4  +  a2&2  +  &4).      10-   (2a?-3y)  (4aj!  +  6*y  +  9jr!). 

These  products  show  that  the  difference  of  the  cubes  of  two 
numbers  is  the  product  of  two  factors,  one  of  which  is  the 
difference  of  the  numbers,  and  the  other  the  sum  of  the  squares 
of  the  numbers  plus  their  product. 

E.g.   (1)  xs  -  ya  =  (x  -  y)(x2  +  xy  +  if). 

(2)  8  a3  -  64  b*  =  (2  a)«  -  (4  6)8 

=  (2a  -46)(4a2-2a-45+  16 b2). 

(3)  a9  -  b9  =  (a3)3  -  (i3)3  =  (a3  -  i3)(a6  +  «363  +  &«). 

Notice  the  difference  between  the  factor  x2+xy-\-y2,  and  the  trino- 
mial square  x2+2  xy+y2. 

EXERCISES 

Determine  whether  each  of  the  following  is  the  difference  of 
two  cubes,  and  if  so,  find  the  factors ;  if  not,  make  it  so  by 
changing  one  of  its  terms.  Check  the  results  by  multi- 
plication. 

1.  a8  — 6s.  8.  64  or*-?/3. 

2.  8a3-&3.  9.  27 -125  a3. 

3.  a3-8  63.  10.  x6-y6. 

4.  8  a3  -  8  K,  11.  1  -  x6. 

5.  33-23.  12.  a9 -8. 

6.  1-a10.  13.  1-125^. 
7.1-8  a3.  14.  8-27  Xs. 

22.  Also  factor  10,  11,  19,  and  20  as  the  difference  of  two 
squares. 


15. 

27^-64. 

16. 

2*s"-l. 

17. 

8  x4  -  f. 

18. 

64  a3-  27  bs. 

19. 

1-729  a". 

20. 

x6-y12. 

21. 

27  r3- 125  s3. 

188  SPECIAL  PRODUCTS  AND  FACTORS 

TRINOMIALS   OF  THE   FORM  x2  +(a  +  6)  x+  ab. 
139.   In  §§  82  to  85  were  found  such  products  as 

(1)  (x  +  5)(x  +  2)=  x2  +  Ix  +  10. 

(2)  (x  -  5)(x  -  2)=  x2  -  7 x  +  10. 

(3)  (a?  +  5)(oj  -  2)=  a,-2  +  Bx  -  10. 

(4)  (a;  -  B)(x  +  2)=  ar>  -  3a;  - 10. 

In  each  case  the  binomials  to  be  multiplied  have  one  term  x 
in  common,  and  the  other  two  terms  unlike.  The  trinomial 
product  in  each  case  is  the  square  of  the  term  common  to  the 
binomials,  the  algebraic  sum  of  the  unlike  terms  times  the 
common  term,  and  the  product  of  the  unlike  terms. 

Thus,  the  coefficient  of  x  in  (1)  is  5  +  2,  in  (2)  it  is  (—  5)  +  (—  2), 
in  (3)  itis(+  5)  +  (-  2),  and  in  (4)  it  is  (-5)+(+  2). 

The  third  term  of  the  product  in  (1)  is  (+  5)(+  2),  in  (2)  it  is 
(-  5)(-  2),  in  (3)  it  is  (+  5)(-  2),  and  in  (4)  it  is  (-  5)(+  2). 

With  these  points  clearly  in  mind,  such  products  may  be 
written  out  at  once,  without  the  formal  work  of  multiplication. 

EXERCISES 

Find  the  following  products  by  inspection : 

1.  (a +  7) (a +9).  9.  (a?-  l)(a?-+ll). 

2.  (6-5)(6  +  7).  10.  (c-3)(c  +  12). 

3.  (x -7) (x -17).  11.  (e»-l)(c»- 5). 

4.  (y  +  8)0  -  3).  12.  (x*  -  5) {x*  -  3). 

5-  (V  +  7)(y  -  9).  13.  (a2  -  2)(a2  -  4). 

6-  (y-7)(y-l).  14.  (a3-l)(a3  +  4). 

7.  (*-3)(aj-4).  15.  (2a-l)(2a-3). 

8.  (x-S)(x-9).  16.  (2a  +  3)(2a-4). 


THE  TYPE  x2  +  (a  +  b)x  +  ab  189 

17.  (x-5)(x  +  2).  25.  (3&c  +  4)(3  6c-7). 

18.  (3  a -7)  (3  a +  2).  26.  (100 +  3) (100 +  5). 

19.  (a2-8)(a2-4).  2?  (2  a«  +  &)(2  a«  +  c). 

20.  (&3-3)(63-4). 

21.  (2a?  +  a)(2a;+6).  V'         A         ' 

22.  (5c-4)(5c  +  5).  29'  (*  +  «)<*-»)• 

23.  O»y+0)(ajy-7).  30-  t?  r- «)(»-. &)• 

24.  (l-3a)(l-2a).  31.  (3x*  +  2  2/)(3^-52/). 

140.  It  is  possible  to  recognize  such  products  at  sight,  and 
thus  to  find  the  factors  by  inspection. 

Illustrative  Examples.  Determine  whether  the  following  tri- 
nomials are  of  the  kind  just  considered : 

1.  x2  +  7  x  + 12.  The  question  is  whether  two  numbers  can- 
be  found  such  that  their  sum  is  +  7  and  their  product  12. 
The  numbers  3  and  4  answer  these  conditions.     Hence, 

x2  +  7  x  + 12  =  (a;  +  3)  (x  +  4). 

2.  x2  —  5  x  — 14.  Since  the  product  of  the  numbers  sought  is 
— 14,  one  number  must  have  the  sign  —  and  the  other  -{-;  and 
since  their  sum  is  —5,  the  one  having  the  greater  absolute 
value  must  have  the  sign  — .  Hence,  the  numbers  are  —  7  and 
+  2,  and  we  have  x2  —  5x—  14=(#  —  7)(x  +  2). 

3.  x2 -7 x  +  12  =  (x-3)(x -4).  Since  (-3)(-4)=  +  12 
and  (-3) +  (-4)  =  -7. 

4.  x2  +  4:X-12  =  (x  +  6)(x-2).  Since  (+6)(-2)=-12 
and(+6)  +  (-2)  =  +  4. 

It  is  to  be  noted  that  it  is  not  always  possible  to  find  inte- 
gers to  fulfill  these  two  conditions. 

E.g.  given  x2  +  5x  +  3.  By  inspection,  it  is  easily  seen  that  there 
are  no  two  integers  such  that  their  sum  is  +  5  and  their  product  -f  3. 


190  SPECIAL   PRODUCTS  AND  FACTORS 

EXERCISES 

Determine  whether  each  of  the  following  trinomials  can  be 
factored  by  inspection,  and  if  so,  find  the  factors ;  if  not,  mod- 
ify one  term  so  as  to  make  such  factoring  possible. 

1.  ^  +  3^  +  2.  11.  b2  +  8b  +  15.  21.  z4  +  18ar°  +  77. 

2.  tf  +  x-6.  12.  b2-b-56.  22.  a4-5a2-104. 

3.  ^_a;_6.  13.  b2  +  b-5S.  23.  a2+32a  +  240. 

4.  x2-6x+8.  14.  c2-3c-15.  24.  a4-lla2  +  28. 

5.  a2  +  6a+8.  15.  rf-lSx  +  m.  25.  a4-lla2-60. 

6.  3^-3^  —  8.  16.  x2  +  15a;-54.  26.  a2 -14  a  — 51. 

7.  a,*2 +  2  a -8.  17.  0^-14  #  —  95.  27.  a2-3a-54. 

8.  a2_4a_32.  18.  y2  +  21  y  +  98.  28.  »4-8a^-32. 

9.  a2  +  4a-32.  19.  y2-7y-98.  29.  a6-3a3-154. 
10.  62  +  15  6  +  56.  20.  a2 -19  a; +  78.  30.  a?-10x  +  25. 

31.  a266  - 13  a&3  -  30.  41.  9  a2 +  24  a +  16. 

32.  x2  - 17  xyz  + 72 y2z2.  42.  81  a2 -99  a +  30. 

33.  r8  +  6  r4s  -  91  s2.  43.  #2  + 26^  +  133. 

34.  a4c4  +  9a2c2-162.  44.  x2  +  5xy-8±y\ 

35.  a2  +  lla-210.  45.  *+3r— 154. 

36.  m4  +  4mV  +  4w2.  46.  u2  +  38  uv  + 165  v2. 

37.  sV-Wst-te.  47.  (a  +  6)2-19<a  +  &)  +  88. 

38.  a2b2  -  27  ab  +  26.  48.  (z-?/)2-14(a;-y)  +  40. 

39.  Z2  +  13Z  +  42.  49.  (r-s)2-17(r-s)  +  60. 

40.  «Y  — 11  xy  — 180.  50.  x2  +  (a  +  6)  x  +  a&. 


THE  TYPE  atf  +  bx  +  c  191 

141.   Trinomials  of  the  form  ax*  +  bx  +  c. 

Ex.     Find  the  product  of  2  x  +  5  and  3  x  +  2. 

2z  +  5 
3a  +  2 

6^  +  15* 

4z+10 
6a2  +  19a  +  10 

The  products  3  a;  •  2#  and  2  •  5  are  called  end  products  and 
2  •  2  a;  and  5  •  3  x  are  called  cross  products.  In  this  case  the 
cross  products  are  similar  with  respect  to  x  and  are  added. 
Hence  the  final  product  is  a  trinomial  two  of  whose  terms  are 
the  end  products  and  the  third  term  is  the  sum  of  the  cross 
products. 

This  fact  enables  us  to  write  such  products  at  once. 

Kg.  (3a  +  4)(5a-7)=15cr-a-28. 

In  this  case  15  a2  is  the  first  end  product  and  —  28  the  second,  while 
—  a  is  the  sum  of  the  two  cross  products,  20  a  and  —  21  a. 

In  this  manner  obtain  the  following  products  : 

1.  (2a  +  3)(a  +  3).  11.  (3<-5)(*  +  4). 

2.  (4o-l)(3a  +  2).  12.  (5x-y)(2x  +  3y). 

3.  (2 a +  5) (a -7).        '  13.  (3x-2y)(x  +  3y). 

4.  (7r  +  8)(3r-6).  14.  (4a-3y)(a+jf). 

5.  (2 x  +  8) (9 x -4).  15.  (3r-2s)(2r  +  s). 

6.  (3  m  — l)(4m  +  3).  16.  (5m-n)(2m  +  »). 

7.  (5s-7)(2s-4).  17.  (5a  +  3x)(3a-±x). 

8.  (2aj-l)(7*  +  4).  18.  (4a-56)(a  +  3&). 

9.  (4n-9)(5w-7).  19.  (3a  +  5  6)(a-6). 
10.  (&y-l)(5y  +  ll).  20.  (3  c-  7  d)(2  c  +  3d). 


192  SPECIAL  PRODUCTS  AND  FACTORS 

142.  Trinomials  in  the  form  of  the  above  products  may  some- 
times be  factored  by  inspection. 

Ex.  1.     Factor :  5  x?  + 16  x  +  3. 

If  this  is  the  product  of  two  binomials  they  must  be  such  that  the 
end  products  are  5x2  and  3  and  the  sum  of  the  cross  products  16  x. 

One  pair  of  binomials  having  the  required  end  products  is  5x  +  3 
and  x  +  1,  another  is  5  x  —  1  and  x  —  3,  and  still  another  5  x  +  1  and 
ar  +  3. 

The  sum  of  the  cross  products  in  the  first  pair  is  8  x,  in  the  second 
pair  —  16x,  and  in  the  third  pair  16  x.  Since  the  sum  of  the  cross 
products  in  the  last  pair  is  the  one  required,  the  factors  are  5x+l 
and  x  +  3. 

Ex.  2.     Factor:  6  a2- 5  a -4. 

In  this  case  as  in  the  one  preceding  there  are  several  pairs  of  bi- 
nomials whose  end  products  are  6  a2  and  —4,  such  as  2  a  — 2  and  3  a  +  2, 
6 a—  1  and  a  +  4,  etc.  By  trial  we  find  that  among  these  3 a  —  4  and 
2  a  + 1  is  the  only  pair  the  sum  of  whose  cross  products  is  —  5  a. 
Hence  6  a2-  5  a  -  4  =  (3  a  -  4)(2  a  +  1). 

Obviously  not  every  trinomial  of  this  form  can  be  factored  in  this 
manner.  Thus,  for  example,  in  6  a2  +  5  a  +  4,  no  pair  of  binomials 
whose  end  products  are  6  a2  and  4  has  the  sum  of  its  cross  products  5  a. 

In  this  manner  factor  the  following : 

1.  3^+5^  +  2.  9.  5r*+18r-& 

2.  9a2  +  9a  +  2.  10.  14 a-2 - 39 a  + 10. 

3.  2** +11* +12.  11.  5x2  +  26a-24. 

4.  9z2  +  36z  +  32.  12.  2a2-5a  +  2. 

5.  2 a* -a; -28.  13.  2rft2-m-3. 

6.  12.s2  +  lls  +  2.  14.  7c2-3c-4. 

7.  6t2  +  7t-3.  15.  5a;4  +  9^-18. 

8.  6^-3-2.  16.  7a4  +  123a2-54. 


FACTORS  FOUND  BY  GROUPING  193 

17.  6c2 -19c +  15.  24.  6-5a;-4iB2. 

18.  3a2-21a  +  30.  25.  3h2-13h  +  U. 

19.  6^  +  4^-2.  26.  lor2-r-2. 

20.  20a2-a-99.  27.  2t2  +  llt  +  5. 

21.  12c2  +  25c  +  12.  28.  10-5X-15X2. 

22.  8  +  6a-5a2.  29.  5 x2 - 33 x  + 18. 

23.  15-507-10^.  30.  20-9a-20ar!. 

FACTORS  FOUND  BY  GROUPING 
143.  Besides  the  methods  of  factoring  which  have  been 
applied  to  the  types  of  expressions  thus  far  considered,  there 
are  various  other  processes  which  will  be  considered  in  the 
Advanced  Course.  One  other  method  of  general  application 
will  suffice  here. 

Ex.  1.     Find  the  factors  of  ax  +  ay  -f-  bx  +  by. 

By  Principle  I,  the  first  two  terms  may  be  added  and  also 
the  last  two. 

Thus,  ax  +  ay  +  bx  +  by  =  a(x  +  y)  +  b(x  +  y). 

The  given  expression  has  thus  been  changed  into  the  sum  of 
two  compound  terms  which  have  a  common  factor,  (x  +  y),  and 
may  be  added  with  respect  to  this  factor  by  Principle  I. 

Thus,  a{x  +  y)  +  b(x  +  y)  =  (a  +  b)(x+y). 

Hence,  ax -{-ay  +  bx-\-by  =  (a-\-b)(x-\-y). 

Ex.  2.     Factor  ax  —  ay  —  bx  +  by. 

Combining  the  first  two  terms  with  respect  to  a  and  the 
second  two  with  respect  to  —  b,  we  have, 

ax  —  ay  —  bx  +  by  =  a(x  —  y)  —  6  (x—y). 


194  SPECIAL  PRODUCTS  AND  FACTORS 

Again  combining  with  respect  to  the  factor  x  —  y, 
ax—ay  —  bx-\-by=(a  —  b)(x  —y). 

The  success  of  this  method  depends  upon  the  possibility  of 
so  grouping  and  combining  the  terms  as  to  reveal  a  common 
compound  term,  §  77.  The  order  of  the  terms  in  the  given 
polynomial  may  be  changed  as  found  desirable. 

EXERCISES 

Factor  the  following  expressions  by  means  of  Principles  I 
and  II : 

1.  aft2  +  ac2  —  db2  —  dc2.  11.    2  n2  —  en  +  2nd  —  cd. 

2.  6  ms— 15  nt  +  9  ns  — 10  mt.  12.   5  ax  — 15  ay  —  3  bx  +  9  by. 

3.  8  ax  — 10  ay  +  4  bx  —  5  by.    13.   3xa — 12  xc  —  a  +  4  c. 

4.  2  a2+  3  a&  —  14  an  —  21  wfc.  14.   3  xy  —  4  wm  -f-  6  my  —  2 xn. 

5.  ac  +  6c  +  ad  +  6d*.  15.   1  mn  +  1  mr  —  2n2  —  2nr. 

6.  ax2  —  ftcc2  —  ay2  +  6y2.  16.    a  —  1  +  a3  —  a2. 

7.  8  ac  -  20  ad-  6  6c  +15  bd.  17.    3s  +  2  +  6s4 +4S3. 

8.  2ax  —  Q>bx  +  Zby—ay.         18.   as2  —  3  bst  —  ast  +  3  bt2. 

9.  5  +  4  a  — 15  c  — 12  ac.  19.   3  m?i  +  6m2  —  2  am  —  an. 
10.    156-6-206c  +  8c.             20.   2ar  +  2as  +  26r  +  26s. 

MISCELLANEOUS  EXERCISES 

Classify  the  following  expressions  according  to  the  fore- 
going types  for  factoring,  and  indicate  the  factors : 

l.<o?  +  5z+6.  5.  4:X2  +  9y2-12xy. 

2.  1  —  Xs.  6.  5  ar>  +  4aa;  +  7  xy. 

3.  a?  + 11  ai  +  30.  7.  2w2  — 6wc-3wy  +  9cy. 

4.  4arJ  +  9y2  +  12:ry.  8.  4x-2-#2. 


MISCELLANEOUS  FACTORING  195 

9.  a3  +  63.  35.  3(a+l)3+4(a+l)2+a  +  l. 

10.  9x2  +  yi  +  Qxy2.  36.  (x  +  a)2  -  (x—  a)2. 

11.  2y2as  +  4:ya2-8ya.  37.  15  m2  +  224  m  - 15. 

12.  ^  +  7^+6.  38.  3.^4-  27  a +42. 

13.  9  0^  +  36^ +  36  a*/2.  39.  cc4  +  49  a2  + 14 aa2. 

14.  9y-9z-2xy  +  2xz.  40.  27  a6 -aV. 

15.  a'-l.  41.  S^  +  aV. 

16.  a4  +  64  +  2a262.  42.  8M-406e  +  3cd-15ce. 

17.  a4 -25.  43.  a2 -11  a +  30. 

18.  27  a3 -125.  44.  0-+2)2-4(a-2)2. 

19.  4a2  +  4a6  +  a&2.  45.  x*  +  9y2-6yx?. 

20.  4  a2  +  9  a4- 12  ax2.  46.  4a2-7ca2- 4d2+ 7cd*2. 

21.  1  +  ar5.  47.  a2  +  15a -16. 

22.  2x2  +  5a  +  3.  48.  18-27c  +  166-24c6. 

23.  36 +  4  a6 +24  or5.  49.  4-(a2  +  &2-2a&)2. 

24.  0_1)2_0  +  1)2.  50.  10r  +  36s-66r-5s. 

25.  8 +  64  a6.  51.  25 +  64^  + 80a3. 

26.  a,c  —  ax  —  4&c  +  4&c.  52.  1000  —  Xs. 

27.  27 -216  a3.  53.  103+ ar*. 

28.  33  +  63a3.  54.  8a3  +  a3&3  +  &2a2. 

29.  25(x  +  l)2-4.  55.  100  -49a4. 

30.  5cx-10c  +  4:dx  —  8d.  56.  100  +  625  +  500. 

31.  4t{x+2)2+y2  +  ±{x  +  2)y.   57.  a2 -17a +  72. 

32.  ra  +  2  r/t  -  5  sa- 10  sA.  58.  a2  +  17 a  +  72. 

33.  -2a26  +  a4  +  62.  59.  a2+1662-8a6. 

34.  2/ia  —  hb+  6 a- 36.  60.  xs  —  y9. 


196  SPECIAL   PRODUCTS  AND  FACTORS 

61.  24  a2 +  37  a -72.  87.  24  a4  c4  +  a6  +  144  c8 a2. 

62.  (8*  + 15s2  — 100.  88.  9s2  +  4?/4-12s?/2-16. 

63.  9«  +  83.  89.  81  +  100s8-  180 s4. 

64.  9s4  +  16?/2  +  24sy  90.  a4  +  27a2  +  180. 

65.  1-1000.  91.  a4  +  3a2 -180. 

66.  16a262+24a&  +  3663.  92.  a4 -3  a2 -180. 

67.  64  +  8.  93.  144-(a4  +  fc2-2a26). 

68.  16a262  +  9aV  +  24a26c.  94.  81  a2 64  +  49 c2- 126 ab2c. 

69.  a2  +  462  +  4a&-4s2.  95.  12 s2 - 23 st  + 10 t2. 

70.  asb6  +  c?.  96.  36  s4  +  12  s2?/4 +  ?/8. 

71.  5^  + 10s3?/2 +  30 or5?/4.  97.  16s6  +  9y4  + 24sy~49. 

72.  16aV  +  4c2s2+16ac2s.  98.  y2  +  35y  +  300. 

73.  a6y3-z\  99.  5  ?/2 -80?/  + 300. 

74.  s4- 7s2- 120.  100.  100  -  (16s2 +  ?/6- 8s?/3). 

75.  9a4&2  — 12a3&  +  4a2.  101.  ac—bc  +  ad—bd. 

76.  8ao  +  27a&7.  102.  6 rd -15 re +22 cd-55ce. 

77.  s4  +  4s2  +  4  — s6.  103.  z3  +  ya  —  ifz3  —  ay*. 

78.  l-125a366.  104.  s4  +  2s2  +  l-s2. 

79.  16  +  16a6+4a262.  105.  60  s2  +  7  s?/ -  ?/2. 

80.  64a3  +  8a2&3.  106.  s2  -  20  s?/ +  75  ?/2. 

81.  36a264  +  o264  + 12 ab*c.  107.  s2- 17s -60. 

82.  4a2  +  9&4  +  12a62-16a4.  108.  65^  +  8  r-1. 

83.  s6  +  17s3  +  30.  109.  a2 -13a -140. 

84.  25-(a4-2a2Z>3  +  &fi).  110.  39s4-16s2  +  l. 

85.  -112a2c3  +  49a4  +  64c6.  111.  625-  (31  -4a2)2. 

86.  a2_a_380.  112.  36a2-29a&  +  5  62. 


EQUATIONS  SOLVED  BY  FACTORING  197 

113.  c4-  31c2  +  220.  115.   26  +  39w-22m-33mw. 

114.  ac  +  d5a—  b4c-b*d5.  116.    12  a2  +  11  a  -  56. 

117.  a2  +  4a&+462-(a2-4a&  +  4&2). 

118.  (3a;-l)2-(a,'2  +  4^-4^)2. 

119.  (x  +  3yy+(x-2yy  +2(x  +  3y)(x-2y). 

120.  16 (a  +  b)2  -  8 (a  -6)  (a  +  6)  +  (a  -  b)\ 

121.  256a,-2  -  (49  x2  +  4?/4  -  28a?/2). 

122.  (2a -a)2 +100  (a -3a)2  +20  (2a  -  a)  (a  -  3a). 

123.  -48(a-6)(a+6)  +  36(a-&)2  +  16(a+6)2. 

EQUATIONS  SOLVED  BY  FACTORING 

144.  Illustrative  Problem.  There  are  two  consecutive  num- 
bers the  sum  of  whose  squares  is  61.     What  are  the  numbers  ? 

Solution.    Let  x  =  one  of  the  numbers,  then  x  +  1  is  the  other. 
Hence,  x2  +  (x  +  l)2  =  61  (1) 

By  F,  x2  +  x*  +  2x  +  1  =  61  (2) 

By  F,  I,  S,  2x2  +  2x        =60  (3) 

By  D,  x2  +  x  =  30  (4) 

These  equations  differ  from  any  which  we  have  studied  here- 
tofore in  that  they  contain  the  squares  of  the  unknown  num- 
ber, which  cannot  be  removed  by  addition  or  subtraction. 

It  is  evident  on  inspection  that  a  =  5  satisfies  equation  (4). 

That  is,  52  +  5  =  30. 

Hence  5  is  one  of  the  numbers  sought,  and  5  +  1  is  the  other,  and 
these  numbers  satisfy  the  conditions  of  the  problem, 

Since  52  +  62  =  25  +  36  =  61. 

145.  Definition.  Equations  which  involve  the  second  but 
no  higher  degree  of  the  unknown  number  are  called  quadratic 
equations. 


198  SPECIAL  PRODUCTS  AND  FACTORS 

One  method  of  solving  quadratic  equations  is  now  to  be 
considered. 

By  S,  equation  (4)  above  may  be  written 

x2  +  x  -  30  =  0.  (5) 

Factoring  the  left  member, 

(x  +  6)  (x  -  5)  =  0.  (6) 

Equation  (6)  is  satisfied  by  any  value  of  x  which,  substituted  in 
the  left  member,  reduces  it  to  zero  ;  and,  since  the  product  of  two  fac- 
tors is  zero  if  either  factor  is  zero,  we  seek  values  of  x  which  make 
x  —  5  =  0  and  also  x  +  6  =  0.  Hence  the  equation  is  satisfied  by 
x  =  5  and  also  by  x  =  —  6.     Thus, 

(5  +  6)  (5  -  5)  =  11  •  0  ss  0, 
and  also  (-  6  +  6)  (-  6  -  5)  =  0  .  -  11    =0. 

Therefore  —  6  and  —  6  +  1  =  —  5  are  two  numbers  which  meet 
the  condition  of  the  problem. 

That  is,  (-  6)2  +  (  -  5)2  =  36  +  25  =  61. 

Hence  this  problem  has  two  solutions,  namely,  the  numbers  5  and  6 
and  the  numbers  —  6  and  —  5. 

Illustrative  Problem.  A  rectangular  flower  bed  10  feet  long 
and  6  feet  wide  is  surrounded  by  a  gravel  walk  whose  area  is 
192  square  feet.     How  wide  is  the  walk  ? 

Solution.    Let  the  width  of  the  walk  be  x  feet,  then 

2  •  10  x  is  the  area  of  the  sides, 
2  •  6  x  is  the  area  of  the  ends, 
4  x2  is  the  area  of  the  corners. 

Hence  the  total  area  is 

4x2  +  2-10x  +  2.6x  =  192.  (1) 

By  F,  4x2+32x  =  192.  (2) 

ByZ>,  x2  +  8x=48.  (3) 


EQUATIONS  SOLVED  BY  FACTORING  199 

By  S,  x2  +  8  x  -  48  =  0.  (4) 

Factoring,  (x  +  12)  (x  -  4)  =  0.  (5) 

But  (5)  is  satisfied  if  x  +  12  =  0,  and  also  if  x  —  4  =  0.  (6) 

Hence  x  =  —  12  and  also  x  =  4.  (7) 

Check  by  substituting  in  equation  (3)  : 

(  -  12)2  +  8  (-  12)  =  144  -  96  a  48. 
Also  42  +  8  •  4  =  16  +  32  =  48. 

Hence  the  quadratic  equation  to  which  this  problem  gives  rise  has 
two  solutions,  but  since  the  width  of  a  walk  cannot  be  a  negative 
number,  only  the  number  4  satisfies  the  conditions  of  the  problem. 

Ex.  1.     Solve  the  quadratic  equation 

5a2  +  30z  +  3  =  3-5».  (1) 

By  A,  S,                          5x2  +  35x  =  3-3  =  0.  (2) 

Factoring,                        5  x  (x  +  7)  =  0.  (3) 

But  (3)  is  satisfied  if  5  x  =  0,  and  also  if  x  +  7  =  0.  (4) 

Hence                                    x  =  0,  and  also  x  =  —  7.  (5) 
Check.     Substitute  x  =  0  and  also  x  =  —  7  in  (1). 

Ex.  2.     Solve  the  quadratic  equation 

6x2  +  llx  =  10.  (1) 

By  S,  6x2+llx-10  =  0.  (2) 

Factoring,  (3  x  -  2)  (2x  +  5)  =  0.  (3) 

But  (3)  is  satisfied  if  3  x  —  2  =  0  and  also  if  2  x  +  5  =  0. 
Hence  x  =  §  and  x  =  —  |. 

Check  by  substituting  each  of  these  values  in  (1). 


200  SPECIAL   PRODUCTS  AND  FACTORS 

Ex.  3.     Solve  the  quadratic  equation 

3z2-5a;-7  =  2a:2-a;-ll.  (1) 

By  .4andS,  x2-4x  +  4  =  0.  (2) 

Factoring,  (x  -  2)  (x  -  2)  =  0.  (3) 

It  follows  from  the  first  factor  and  also  from  the  second  that  (3)  is 
satisfied  if  x  =  2. 

In  this  case  the  two  solutions  turn  out  to  be  identical,  while  in 
Example  1  one  solution  was  zero  and  the  other  was  —  7. 

If  we  count  each  of  these  results  as  two  solutions,  then  for  every 
quadratic  equation  thus  far  solved  we  have  found  two  values  of  the 
unknown  number. 

146.  The  method  of  solution  above  explained  consists  of 
three  steps : 

(1)  Transform  the  equation  so  that  all  terms  are  collected 
in  one  member,  with  similar  terms  united,  leaving  the  other 
member  zero.  This  can  always  be  done  by  Principle  VIII. 
It  is  convenient  to  make  the  right  member  zero. 

(2)  Factor  the  expression  on  the  left. 

(3)  Find  the  value  of  x  which  makes  each  of  these  factors 
zero.  This  is  readily  done  by  setting  each  factor  equal  to 
zero  and  solving  it  for  the  unknown. 

EXERCISES 

Find  two  solutions  for  each  of  the  following  quadratic 
equations : 

1.  ar,-3»  +  2  =  0.  6.  a2  +  3a  =  10 a  +  18. 

2.  x,  +  7«  =  30.  7.  a2  +  10a=-24-4a. 

3.  a2-lla=-30.  8.  2z2-6:r=-40+12a;. 

4.  a2 +  13 a  =  30.  9.  3x  +  x2=  20^-72. 

5.  a2  +  10a  +  8  =  -3a-34.  10.  17a  +  30= -^-40. 


EQUATIONS   SOLVED  BY  FACTORING  201 

11.  7aj»  +  2a;  =  30a>-21.  21.  x2  +  12  a  +  6  =  5a-4. 

12.  lla  +  3x2  =  20.  22.  2a2-7a=60  +  7tf. 

13.  l§-5x+x2=  -2x*-20x-2.    23.  60  a; +  4  ^  +  144  =  8  a. 

14.  or* -16  =  0.  24.  18  se  =  63  -  x2. 

15.  ^-1=0.  25.  24x2  =  12a  +  12. 

16.  x2  —  x  =  0.  26.  2  a;  =  63—  x2. 

17.  x2  +  a-  =  0.  27.  22  x  +  x2  =  363. 

18.  4  a2  =  25.  28.  3a^  +  7a;  =  6. 

19.  arJ  +  4a;  +  4  =  0.  29.  2^  =  2- 3». 

20.  arJ  +  8a;  +  16  =  0.  30.  x-2  =  -3x2. 

147.  It  is  sometimes  possible  to  solve  other  equations  than 
quadratics  by  the  above  process. 

Ex.  1.    Solve  the  equation  : 

a8 +  30  x  =  11  x2.  (1) 

By  S,  x»-llx2+30x  =  0.  (2) 

By  §131,  x(x2-llx  +  30)  =  0.  (3) 

B\  §  139,  x  (x  -  5)  (x  -  6)  =  0.  (4) 

(4)  is  satisfied  if  x  =  0,  if  x  —  5  =  0,  and  if  x  —  6  =  0. 
Hence  the  solutions  are  x  =  0,  x  =  5,  x  =  6. 

Ex.  2.    a>(»  +  l)(a;-2)(a;+3)  =  0. 

Any  value  of  x  which  makes  one  of  these  factors  zero  re- 
duces the  product  to  zero  and  hence  satisfies  the  equation. 
Hence  the  solutions  of  the  equation  are  found  from 

x  =  0,  a; +  1=0,  x  —  2  =  0,  and  #  + 3  =  0. 

That  is  x  =  0,  x=  —  1,  x  =  2,  x=  —  3  are  the  values  of  x 
which  satisfy  the  equation. 

Notice  that  this  process  is  applicable  only  when  one  member 
of  the  equation  is  zero  and  the  other  member  is  factored. 


202  SPECIAL   PRODUCTS  AND  FACTORS 

EXERCISES 

Solve  the  following  equations  by  factoring : 

1.  x3  —  x2  =  6x.  8.  x2  —  ax  —  bx  +  ab  =  0. 

2.  5a;  =  4ar!  +  arJ.  9.  4 (x _ 2)2 _  (a?  +  3)2  =  0. 

3.  Xs  —  25  a  =  0.  10.  ar'-aic  +  fo;  —  ab  =  0. 

4.  Xs  — 3x2=— 2 x.  11.  x2  +  ax  —  bx  —  ab  =  0. 

5.  3  ^  =  15  or* +  42  a.  12.  9(>  +  2)2-4  (z-3)2  =  0. 

6.  5  a3 +  315  a  =  80  a2.  13.  ^-aj-3^  +  3  =  0. 

7.  a*  +  ax  +  &a;  +  a&  =  0.  14.  or5 -4a;- 8  a2 +  32  =  0. 

15.  5  ^  +  120  a2  =  119  a  -2a^  +  8a^. 

16.  (a?+6aj-16)(a>*-7a;  +  12)  =  0. 

17.  (aj  +  7)(a2-T7a;  +  72)  =  0. 

18.  (a?-16)(«8  +  llaj8+30aj)  =  0. 

148.  Illustrative  Problem.  The  paving  of  a  square  court 
costs  40^  per  square  yard  and  the  fence  around  it  costs  $  1.50 
per  linear  yard.  If  the  total  cost  of  the  pavement  and  the 
fence  is  $  100,  what  is  the  size  of  the  court  ? 

Solution.     Let  x  —  the  length  of  one  side  in  yards. 
Then  40  x*  =  cost  in  cents  of  paving  the  court, 

150  •  4  x  =  600  x  =  cost  of  the  fence  in  cents. 

40  x2  +  600  x  =  10000.  (1) 

By  A  x2+  15  a;  =  250.  (2) 

By  S,  x*  +  15s  -  250  =  0.  (3) 

Factoring,  (x  -  10)  (x  +  25)  =  0.  (4) 

Whence,  x  =  10,  and  also  x  =  —  25.  (5) 

It  is  clear  that  the  length  of  a  side  of  the  court  cannot  be  —  25 
yards.  Hence  10  is  the  only  one  of  these  two  results  which  has  a 
meaning  in  this  problem. 

It  happens  frequently  when  a  quadratic  equation  is  used  to 
solve  a  problem  that  one  of  the  two  numbers  which  satisfy  this 
equation  will  not  satisfy  the  conditions  of  the  problem. 


EQUATIONS  SOLVED  BY  FACTORING 


203 


PROBLEMS 

In  each  of  the  following  problems  find  all  the  solutions  pos- 
sible for  the  equations  and  then  determine  whether  or  not  each 
solution  has  a  reasonable  interpretation  in  the  problem. 

1.  The  dimensions  of  a  picture  inside  the  frame  are  12  by  16 
inches.  What  is  the  width  of  the  frame  if  its  area  is  288 
square  inches  ? 

2.  There  are  two  consecutive  integers  such  that  the  sum  of 
their  squares  is  3961.     What  are  the  numbers  ? 

3.  An  open  box  is  made  from  a  square  piece  of  tin  by  cutting 
out  a  5-inch  square  from  each  corner  and  turning  up  the  sides. 
How  large  is  the  original  square  if  the  box 
contains  180  cubic  inches  ? 

If  x  —  length  of  a  side  of  the  tin,  then  the  volume  of 
the  box  is :  5(x  - 10)  (x  - 10)  =  180.    (See  the  figure.) 

4.  A  rectangular  piece  of  tin  is  4  inches 
longer  than  it  is  wide.  An  open  box  contain- 
ing 840  cubic  inches  is  made  by  cutting  a  6-inch  square  from 
each  corner  and  turning  up  the  ends  and  sides.  What  are 
the  dimensions  of  the  box  ? 

5.  A  farmer  has  a  rectangular  wheat  field  160  rods  long  by 
80  rods  wide.     In  cutting  the  grain,  he  cuts  a  strip  of  equal 

width  around  the  field. 
How  many  acres  has 
he  cut  when  the 
width  of  the  strip  is 
8  rods  ? 

6.   How  wide  is  the 
strip  around  the  field 
of  problem  5,  if  it  con- 
tains 27^  acres  ? 
7.  In  the  northwest  a  farmer  using  a  steam  plow  starts  plowing 
around  a  rectangular  field  640  by  320  rods.    If  the  strip  plowed 
the  first  day  lacks  16  square  rods  of  being  24  acres,  how  wide  is  it  ? 


160 


80 


X 
1 

8 

X 

*(l60-2a:) 
160-2* 

\x  ** 

X 

H 
M 

8 

160-2* 

f 

8 

.so 


160 


204 


SPECIAL  PRODUCTS  AND  FACTORS 


8.   A  rectangular  piece  of  ground  840  by  640  feet  is  divided 
into  4  city  blocks  by  two  streets  60  feet  wide  running  through 

840  feet  it   at  right   angles.      How   many  square 

feet  are  contained  in  the  streets  ? 

9.  A  farmer  lays  out  two  roads 
through  the  middle  of  his  farm,  one 
running  lengthwise  of  the  farm  and  the 
other    crosswise.      How    wide    are    the 

roads  if  the  farm  is  320  by  240  rods,  and  the  area  occupied 

by  the  roads  is  1671  square  rods  ? 


QUADRATIC   AND   LINEAR  EQUATIONS 

149.  When  two  simultaneous  equations  are  given,  one  quad- 
ratic and  one  linear,  they  may  be  solved  by  the  process  of  sub- 
stitution, which  was  used  (§  116)  in  the  case  of  two  linear 
equations. 

Illustrative  Example.    Solve  the  equations : 
r  x1  —  y'1  =  — 16. 
I  a?  —  3  2/  =  — 12. 

From  (2)  by  S,  x  =  3  y  -  12. 

Substituting  (3)  in  (1),       (3  y  -  12)2  -  y2  =  -  16. 
From  (4)  by  F,  9  y2  -  72  y  +  144  -y2=-  16. 

From  (5)  by  F,  A,  8  y2  -  T2  y  +  160  =  0. 


By  A  y2  -  9  y  +  20  =  0. 

Factoring,  (y  —  5)(y  —  4)  =  0. 

Hence,  y  =  5,  and  y  =  4. 

Substitute  y  =  5  in  (2)  and  find  x  =  3. 

Substitute  y  =  4  in  (2)  and  find  x  =  0. 

Therefore  (1)  and  (2)  are  satisfied  by  the  two  pairs  of  values, 

x  =  3,  y  —  5,  and  x  —  0,  y  —  4. 

Check  by  substituting  these  pairs  of  values  in  (1)  and  (2). 


(1) 
(2) 
(3) 
(4) 
(5) 
(6) 
(7) 
(8) 
(9) 


QUADRATIC  AND  LINEAR  EQUATIONS  205 

EXERCISES 

In  the  manner  just  illustrated  solve  the  following : 
\x+2y  =  8,  \x-y=-l, 


2. 


3. 


5««  +  12?/2  =  128.  Ua;2  +  3i/2  =  147. 

1^  +  ^  =  1.  '    Ix2—  52/2  =  4. 

f  2  a?  —  y  =  6,  13     fas-yssi, 

U^  +  5?/2  =  36.  '    1 3  as2  —  2  2/2  =  —  5. 

fa;  +  3y  =  6,  J5z-7y=-28, 

la2 +  3/ =12.  '    115  a2 +  49 1/2  =  784. 


a;_22/=-2,  j6z-7*/  =  18, 

x2-6?/2  =  10.  "    1 36  ^-7^  =  324. 


.8* -16?/ =  -120,  16    Ja-9y  =  2 


7. 


\x-\ 
7 x*  +  2 y2  =  585.  '    Ick2- 45^  =  4. 

j7a  +  9?/  =  88,  f«  +  y=8, 

17^4-9^  =  736.  '    ll3»2  +  32/2  =  160. 

8     ix-y  =  6,  \2x-5y=-16, 

U2-7?/2=36.  '   14^  +  15/ =256. 

|3a;4-22/  =  7,  J7a  +  4y  =  7, 

l3^  +  8y2=35.  '    l49a2-8?/2  =  49. 

f»-8y«-ll,  fa»-3y^-12, 

*   1 3  cc2- 16/ =  11.  '   lx2-y2  =  -16. 

150.  Illustrative  Problem.  The  fence  around  a  rectangular 
field  is  280  rods  long.  What  are  the  dimensions  of  the  field, 
if  its  area  is  30  acres  ? 

Solution.     Let  w  =  the  width  of  the  field  in  rods, 

and  I  sr  length  of  field  in  rods.  (1) 

Then  2 1  +  2  w  =  280, 

and  lw  —  4800    (one  acre  =  160  square  rods). 

(2) 


206  SPECIAL  PRODUCTS  AND  FACTORS 

From  (1),  I  +  w  =  140,  and  w  =  140  -  I. 

Substituting  this  value  of  w  in  (2), 

Z(140  -  0=  4800, 
or,  Z2  -U01  +  4800  =  0. 

Then,  (I  -  60)  (/  -  80)  =  0. 

Whence  60  and  80  both  satisfy  the  equation. 
If  I  =  60,  then  from  equation  (1)  w  =  80,  and  if  I  =  80,  then  w  =  60. 

We  group  these  pairs  of  numbers  as  follows : 

|  /  =  60>andM  =  80, 
I  w  =  80,  I  w  =  60. 

Substituting  these  pairs  of  values  in  both  (1)  and  (2),  we  have 

|  2  •  60  +    2  •  80  =    280,  j  2  •  80  +    2  •  60  =    280, 

I  60  •  80  =  4800,  an     1  80  •  60  =  4800. 

Hence,  we  obtain  two  pairs  of  numbers  which  satisfy  both  of  these 
equations.  The  solution  I  =  60  and  w  =  80  is  applicable  to  this  prob- 
lem only  by  calling  the  greater  side  the  width  instead  of  the  length. 
AVe  have  seen  before  that  solving  a  quadratic  often  results  in  one 
solution  which  is  without  meaning  in  the  problem  that  gives  rise  to 
the  equation. 

In  any  case  each  solution  should  be  carefully  examined  to  ascertain 
whether  under  any  reasonable  interpretation  they  are  both  applicable 
to  the  problem. 

151.  If  squares  are  constructed  on  the  two  sides,  and  also  on 
the  hypotenuse  of  a  right-angled  triangle,  then  the  sum  of  the 
squares  on  the  sides  is  equal  to  the  square  on  the  hypotenuse. 
This  is  proved  in  geometry,  but  may  be  verified  by  counting 
squares  in  the  accompanying  figure.  This  proposition  was  first 
discovered  by  the  great  philosopher  and  mathematician  Py- 
thagoras, who  lived  about  550  b.c.  Hence  it  is  called  the 
Pythagorean  proposition. 

We  now  proceed  to  solve  some  problems  by  this  proposition. 


QUADRATIC  AND  LINEAR  EQUATIONS 


207 


^ 

0 

s 

t 

1 

11 

rf 

6 

s 

1- 

*k 

f<- 

0 

s 

1- 

ft 

i 

PROBLEMS 

1.  The  sum  of  the  sides  about  the  right  angle  of  a  right  tri- 
angle is  35  inches,  and  the  hypotenuse  is  25  inches.  Find  the 
sides  of  the  triangle. 

Solution.  Let  a  =  the  length  of  one  side  in  inches, 

and  b  =  the  length  of  the  other. 

Then  a  +  b  =  35,                                                                        (1) 

and  a2  +  b2  =  252  =  625   (Pythagorean  proposition).           (2) 


208  SPECIAL   PRODUCTS  AND  FACTORS 

From  (1),  a  =  35  -  ft. 

Substituting  in  (2),  (35  -  ft)2  +  ft2  =  625, 

or  1225  -  70  b  +  ft2  +  ft2  =  625, 

2  ft2  -  70  ft  +  600  =  0, 
ft2 -35ft +  300  =  0, 
(ft-  20) (ft-  15)=  0. 

Whence  ft  =  20,  and  ft  =  15. 

From  (1),  if  ft  =  20,  a  =  15,  and  if  ft  =  15,  a  =  20;  that  is,  the 
sides  of  the  triangle  are  15  and  20. 

2.  The  difference  between  the  two  sides  of  a  right  triangle 
is  2  feet,  and  the  length  of  the  hypotenuse  is  10  feet.  Find 
the  two  sides. 

3.  The  sum  of  the  length  and  width  of  a  rectangle  is 
17  rods,  and  the  diagonal  is  13  rods.  Find  the  dimensions  of 
the  rectangle. 

4.  A  room  is  3  feet  longer  than  it  is  wide,  and  the 
length  of  the  diagonal  is  15  feet.  Find  the  dimensions  of 
the  room. 

5.  The  length  of  the  molding  around  a  rectangular  room 
is  46  feet,  and  the  diagonal  of  the  room  is  17  feet.  Find  its 
dimensions. 

6.  The  longest  rod  that  can  be  placed  flat  on  the  bottom 
of  a  certain  trunk  is  45  inches.  The  trunk  is  9  inches 
longer  than  it  is  wide.  What  are  the  dimensions  of  the 
bottom  ? 

7.  The  floor  space  of  a  rectangular  room  is  180  square  feet, 
and  the  length  of  the  molding  around  the  room  is  56  feet. 
What  are  the  dimensions  of  the  room  ? 

8.  A  rectangular  field  is  20  rods  longer  than  it  is  wide,  and 
its  area  is  2400  square  rods.     What  are  its  dimensions  ? 

9.  A  ceiling  requires  24  square  yards  of  paper,  and  the 
border  is  20  yards  long.  What  are  the  dimensions  of  the 
ceiling  ? 


QUADRATIC  AND   LINEAR  EQUATIONS  209 

10.  The  area  of  a  certain  triangle  is  18  square  inches,  and 
the  sum  of  the  base  and  altitude  is  12.  Find  the  base  and 
altitude. 

11.  The  altitude  of  a  certain  triangle  is  7  inches  less  than 
the  base,  and  the  area  is  130  inches.  Find  the  base  and 
altitude. 

12.  The  sum  of  two  numbers  is  17,  and  the  sum  of  their 
squares  is  145.     Find  the  numbers. 

13.  The  difference  of  two  numbers  is  8,  and  the  sum  of  their 
squares  is  274.     Find  the  numbers. 

14.  The  difference  of  two  numbers  is  13,  and  the  difference 
of  their  squares  is  481.     Find  the  numbers. 

15.  The  sum  of  two  numbers  is  40,  and  the  difference  of 
their  squares  is  320.     Find  the  numbers. 

16.  The  sum  of  two  numbers  is  45,  and  their  product  is 
450.     Find  the  numbers. 

17.  The  difference  of  two  numbers  is  32,  and  their  product 
is  833.     What  are  the  numbers  ? 


CHAPTER   VII 

QUOTIENTS   AND   SQUARE   ROOTS 

QUOTIENT   OF   TWO   POWERS    OF   THE    SAME   BASE 

152.  Illustrative  Problem.    To  divide  x6  by  x\ 

Since  by  §  66  the  quotient  times  the  divisor  equals  the  dividend, 
we  seek  an  expression  which  multiplied  by  x*  equals  x6. 

Since  by  Principle  XIV  two  powers  of  the  same  base  are  multiplied 
by  adding  their  exponents,  the  expression  sought  must  be  that  power 
of  x  whose  exponent  added  to  4  equals  6.  Hence  the  exponent  of  the 
quotient  is  6  -  4  =  2.     That  is,  x*  ■*■  a;4  =  x6-4  =  x2. 

EXERCISES 

Perform  the  following  indicated  divisions  : 

1.  24-=-22.  8.  513-=-512.  15.  x*  +  x*. 

2.  23--22.  9.  T-4--!22.  16.  tu  +  t*. 

3.  24-=-2.  10.  83-=-8.  17.  m3-*-ra. 

4.  33-=-32.  11.  64-=-62.  18.  n6-=-n2. 

5.  34h-3.  12.  as  +  a2.  19.  (20)4h-(20). 

6.  34-5-32.  13.  a*--- a3.  20.  (101)14--(101)13. 

7.  9u-*-9*.  14.  m4-t-m*.  21.  417-;-416. 

153.  The  process  of  division  by  subtracting  exponents 
leads  in  certain  cases  to  strange  results. 

Thus,  according  to  this  process,  x4  +  x4  =  a;4-4  =  x°,  which  is  as  yet 
without  meaning,  since  an  exponent  has  been  defined  only  when  it 
is  a  positive  integer.  It  cannot  indicate,  as  in  the  case  of  a  positive 
integral  exponent,  how  many  times  the  base  is  used  as  a  factor.  We 
know,  however,  that  x4  +  x4  —  1,  since  any  number  divided  by  itself 

210 


QUOTIENTS  OF  POWERS   OF  THE  SAME  BASE     211 

equals  unity.  Hence  if  we  use  the  symbol  x°  it  must  be  interpreted  to 
mean  1,  no  matter  what  number  x  represents.  It  is  sometimes  con- 
venient in  algebraic  work  to  use  it  in  this  way. 

Again  by  this  process  x2  +  x*  =  x2-4  =  x-2,  which  is  as  yet  without 
meaning,  since  negative  exponents  have  not  been  defined.  Cases  of 
this  kind  are  considered  in  the  Advanced  Coui-se. 

The  preceding  exercises  illustrate  the  following  principle : 

154.  Principle  XVI.  The  quotient  of  two  powers  of  the 
same  base  is  a  power  of  that  base  whose  exponent  is  the 
exponent  of  the  dividend  minus  that  of  tJie  divisor. 

For  the  present  only  those  cases  are  considered  in  which  the  expo- 
nent of  the  dividend  is  greater  than  or  equal  to  that  of  the  divisor. 

Notice  that  Principle  XVI  does  not  apply  to  powers  of  different 
bases. 

E.g.  37  +  24  does  not  equal  any  integral  base  to  the  power,  7  —  4. 
This  division  can  be  performed  only  by  first  multiplying  out  both 
dividend  and  divisor. 

EXERCISES 

Perform  the  following  indicated  divisions  by  means  of  Prin- 
ciple XVI : 

1.  27-r-23.  6.  xin  +  x2n.  11.  a?*+h  -+-  x°+b. 

2.  a7  +  as.  7.  32"-1 -f- 3a"2.  12.  ic^  +  w'. 

3.  34-=-32.  8.  5"+5--5"+2.  13.  (17)14h- (17)13. 

4.  x*  +  x*.  9.  xa+i-^xa+2.  14.  43-i-4. 

5.  33"-*-32n.  10.  t4a  +  ta.  15.  (12)4--(12)3. 

In  the  following  use  Principles  V,  VI,  and  XVI : 

16.  (24  +  23) -=- 23.  21.   (a2ra4-&2m3)-r-m3. 

17.  (3  •  24  +  5  •  23)  -=-  22.  22.    (4  •  32  -  33  •  5  •  7)  -f-  32. 

18.  (3.43-5-44)-=-42.  23.   (23  •  3  +  24  •  33  -  23) -23. 

19.  (a3b  -  a462)  -*-  a2.  24.    (12afy-  llaV+5a;4)  +  X2. 

20.  (4  ^  +  3  a4)-*- a?2.  25.    (x3m+4+x2m+3-5xm+2)-i-xm+\ 


212  QUOTIENTS  AND  SQUARE  BOOTS 

DIVISION   OF   MONOMIALS 

155.  In  finding  the  quotient  of  two  numbers  each  in  the 
factored  form,  if  the  factors  are  represented  by  Arabic  figures, 
the  operation  may  be  carried  out  in  either  of  two  ways. 

E.g.  28  •  3» .  5  -  2  •  3  =  1080  -  12  =  90. 

Also  23  •  33  •  5  -  22  •  3  =  2  •  32  •  5  =  90. 

In  the  second  process  we  divide  by  one  of  the  factors,  22  or  3, 
and  divide  this  result  by  the  other.  23  •  33  •  5  divided  by  22 
gives,  according  to  Principle  V,  2  •  33  •  5,  and  this  result  di- 
vided by  3  gives  by  the  same  principle  2  •  32  •  5.  In  practice 
such  operations  may  readily  be  performed  simultaneously. 

In  the  case  of  literal  factors  the  second  process  only  is 
available. 

E.g.  5  aWc  +  a262  =  5  a4-268~2c  =  5  a'Wc  =  5  a2bc. 

EXERCISES 

Perform  the  following  indicated  divisions  in  two  ways  when 
possible : 

1.  58-79-h5-72.  5.    15asb*  +  5ab. 

2.  23-33-53-^2.32-5.  6.   12  x*y  ~  4  x. 

3.  44.5<h_4.53.  7.    18  ^--3  J2. 

4.  34-52-5-3-5.  8.    28m3n-=-7m. 

The  preceding  examples  illustrate  the  following  principle : 

156.  Principle  XVII.  TJie  quotient  of  two  monomials  is 
found  by  dividing  the  dividend  by  each  factor  of  the  divi- 
sor in  succession. 

Each  factor  of  the  divisor  is  associated  with  any  desired 
factor  of  the  dividend  according  to  Principle  V,  and  when  the 
bases  are  the  same  the  exponents  are  subtracted  according  to 
Principle  XVI. 


DIVISION   OF  MONOMIALS  213 

157.  If  there  are  factors  in  the  divisor  not  found  in  the 
dividend,  this  process  terminates  before  the  operation  is  com- 
pleted. The  remaining  steps  of  the  division  must  then  be  indi- 
cated, which  is  usually  done  in  the  form  of  a  fraction. 

E.g.  x2-r-3  xsy  =  — — —  =  — . 

3  x1  •  xy     3  xy 

Again,  15  a%2c  +  3  a2bx2y  =  15^>%  =  £«*? . 

3  a-bx2y       x'2y 

By  this  process  all  factors  common  to  dividend  and  divisor 
have  been  canceled.  Notice  that  in  the  first  example  the  fac- 
tor 1  remains  when  x2  of  the  dividend  is  divided  by  x2  of  the 
divisor. 

_L.     .  ,  EXERCISES 

Divide : 

1.  4  •  7  •  9  by  2  .  3.  6.  5  a4bnc  by  ab*<*. 

2.  12  •  8  •  20  by  2  •  4  •  5.  7.  10  aWV  by  2xb4c. 

3.  6  x?y2z  by  2  xyz.  8.  36  x4f  by  6  xSf . 

4.  64  •  34  •  x3  by  62  •  52  •  x2.  9.  35  a*- y+i  by  5  xa~hf+\ 

5.  12  x12y13  by  4  xtfz.  10.  2  m20*4™3—2  by  ma+2na-2. 

In  each  of  the  following  exercises  state  which  of  the  Prin- 
ciples I-XVII  are  used : 

Divide : 

11.  23.32-24.33by  23 .  32. 

12.  5-27.38  +  7-25.34by  2s  •  34. 

13.  4  aV-  3  x?y2  by  x2y2. 

14.  18  xy  - 12  xAf  +  6  x2y2  by  6  afy2. 

15.  49  a4 -f  21  a3- 7  a  by  la. 

16.  12  ax4}/3  — 16  a2ary  -f  8  a3xy  by  4  axy. 

17.  2  ar3*  +  4  x4"  —  8  x2"  by  2  a8. 

18.  6  x2" f  J  -f  12  ar3n+]  -  10  iC+1  by  2  an+1. 

19.  4  a;13  -  6  «n6  —  10  x4c  by  2  x4. 

20.  10  a362  -  a2b3  + 15  u464  by  5  a2b2. 


214  QUOTIENTS  AND   SQUARE  ROOTS 

SQUARE  ROOTS  OF  MONOMIALS 

158.  Definition.  The  radical  sign,  Vj  indicates  that  we  are  to 
find  one  of  the  two  equal  factors  of  the  number  expression 
which  follows  it,  and  the  vinculum  is  attached  to  it,  V  ,  to 
show  how  far  its  effect  is  to  extend. 

E.g.  V9  is  read  the  square  root  of  9. 
Similarly,  Va2  -f  2  ab  +  b2  is  read  the  square  root  of  a2  +  2  ab  +  b2. 

The  square  root  of  any  number  is  at  once  evident  if  we  can 
resolve  it  into  two  equal  groups  of  factors. 

E-g. 

V576  =  V2- 2- 2- 2- 2 -2.  3. 3=  V(23.  3)(28  .  3)  =  V24  .  24  =  24. 

It  should  be  noted  that  every  square  has  two  square  roots. 
E.g.  V9  =  -  3  as  well  as  +  3,  since  (-  3)2  =  9  and  32  =  9. 

In  obtaining  a  square  root  the  two  results  should  always 
be  indicated.  This  is  usually  done  by  attaching  the  double 
sign  ±  to  the  square  root,  e.g.  V9  =  ±  3. 


EXERCISES 

Find  by  i 

inspection  the  following 

;  square  roots : 

i.  VI. 

8. 

VI21. 

15.    V324. 

22. 

V24". 

2.    V9. 

9. 

VI69. 

16.    V289. 

23. 

V51"2. 

3.    VI6. 

10. 

V225. 

17.    V625. 

24. 

V71". 

4.    V25. 

11. 
12. 

V196. 

V256. 

18.    V900. 

25. 
26. 

Va12. 

5.    V36. 

19.  Vioooo. 

V31"4. 

6.    V49. 

13. 

V576. 

20.    Va4. 

27. 

Va14. 

7.    V8T. 

14. 

VIoo. 

21.    Vi?. 

28. 

V3^. 

SQUARE  BOOTS  OF  MONOMIALS  215 

159.  The  square  root  of  the  product  of  several  factors,  each 
of  which  is  a  square,  may  be  found  in  two  ways  if  the  factors 
are  expressed  in  Arabic  figures. 

E.g.  V4  •  16  •  25  =  V1600  =  V40  .  40  =  ±  40, 

or  V4  •  16  •  25  =  V2'2 .  42  •  52  =  ±  2  •  4  •  5  =  ±  40. 

But  with  literal  factors,  the  second  process  only  is  available. 


E.g.   Vl6  a*b4c*  =  V42  a%*c2  =  ±  4  ab2c. 


EXERCISES 


Find  the  following  indicated  square  roots,  keeping  the  re- 
sults in  the  factored  form  : 


1.    V22-32.  7.    V312-514.  13.    V9zy2. 


2.   V81  •  121.  8.    V222  •  312.  14.    V121  a¥. 


3.    V49-25-169.       9.    Vl6aW.  15.    V74a462. 


4.    V82.52-32.  10.    V64aV.  16.    V625  x*y\ 


5.  V54.32-44.  11.    V44a264.  17.    Vl225a2r. 

6.  V25  •  36.  12.    ^W¥y\  18.    V36r&4™c2". 

Notice  that  the  square  root  of  a  sum  is  not  obtained  by  taking 
the  square  roots  of  the  terms  separately.  Thus,  V9  + 16  is 
not  equal  to  V9  +  Vl6. 

The  preceding  exercises  illustrate  the  following  principle : 

160.  Principle  XVIII.  The  square  root  of  a  product  is 
obtained  by  finding  the  square  root  of  each  factor  sepa- 
rately and  then  taking  the  product  of  these  roots. 

In  order  that  a  factor  may  be  a  perfect  square  it  must  be 
a  power  whose  exponent  is  even.     Its  square  root  is  then  a 


216  QUOTIENTS  AND  SQUARE  ROOTS 

power  of  the  same  base  whose  exponent  is  equal  to  one-half 
the  given  exponent. 

Thus,  Vx^=  Vx3  •  xs  =  xs.  The  exponent  3  of  the  root  is  one-half 
the  exponent  6  of  the  power.  Hence  to  find  the  square  root  of  a  mo- 
nomial we  divide  the  exponent  of  each  factor  by  2. 

EXERCISES 

Find  the  following  square  roots  : 

1.  V4a%«.                   6.    VlO4 a4b\  11.  V81  x*y8cw. 

2.  V3VV4.                 7.    Vo^m".  12.  V729  a«y™zH. 

3.  V5-  •  3-  fM.              8.    V54^38^72.  13.  V64  •  625  a2b\ 

4.  V121  xy\              9.    V314  -  712  a4.  14.  V216  arV". 

5.  V576  a2b*.             10.    V2^W2.  15.  V32*  •  52". 

DIVISION  BY  A   POLYNOMIAL 
161.    The  simplest  case  of  division  by  a  polynomial  is  that 
in  which  the  dividend  can  be  resolved  into  two  factors,  one 
being  the  given  polynomial  divisor  and  the  other  a  monomial. 

E.g.  To  divide  4  x3  +  4  x2y  by  x  +  y,  factor  the  dividend  and  we 
have  4  x2  (x+y)  h-  (x  +  y)  =  4  x2. 

In  case  the  dividend  cannot  be  factored  in  this  manner, 
then,  if  the  division  is  possible,  the  quotient  must  be  a  polyno- 
mial. The  process  of  finding  the  quotient  under  such  circum- 
stances is  best  shown  by  studying  a  particular  case. 

Illustrative  Example  1.     Consider  the  product 

(x2  +  2xy  +  y2)(x+y)  =  x2(x  +  y)  +  2  xy  (x  +y)  +  y2  (x  +  y). 

The  products,  x2(x  +  y),  2xy(x  +  y),  and  y2(x  +  y)  are  called  partial 
products,  and  their  sum,  x3  +  3  x2y  +  3  xy2  +  y3,  the  complete  product. 

In  dividing  x3  +  3  x2y  +  3  xy2  +  y3  by  x  +  y  the  quotient  must  be 
such  a  polynomial  that  when  its  terms  are  multiplied  by  x  -f  y  the 


DIVISION  BY  A   POLYNOMIAL  217 

results  are  these  partial  products,  which  in  the  solution  are  called  1st, 
2d,  and  3d  products. 

The  work  may  be  arranged  as  follows : 


Dividend  or  product :                x3  +  3  x2y  +  3  xy2  +  y3 
1st  product,  x2(x+y)  :  x3  +  x-y 


x  +  y,  divisor. 


x2  +  2  xy  +  y2, 


Dividend  minus  1st  product:  2  x2y  +  3  xy2  -f  y3     [quotient. 

2d  product,  2xy(x  +  y):  2  x'2y  +  2  xy2 

Dividend  minus  1st  and  2d  products :  xy2  +  y3 

3d  product,  y\x+y):  xy2  +  y3 

Dividend  minus  1st,  2d,  and  3d  products  :  0 

Explanation.  Since  the  dividend  or  product  contains  the 
term  x3,  and  since  one  of  the  factors,  the  divisor,  contains  the 
term  x,  the  other  factor,  the  quotient,  must  contain  the  term 
x2.  Multiplying  this  term  of  the  quotient  by  the  divisor,  we 
obtain  the  first  partial  product,  x3  +  x*y. 

Subtracting  the  first  partial  product  from  the  whole  product 
Xs  +  3  x2y  +  3  xy2  +  y3,  the  remainder  is  2  x2y  +  3xy2  +  y3,  which 
is  the  product  of  the  divisor  and  that  part  of  the  quotient 
which  has  not  yet  been  found.  Since  this  product  contains  the 
term  2  x?y  and  the  divisor  contains  the  term  x,  the  quotient 
must  contain  the  term  2  xy.  The  product  of  2  xy  and  x  +  y  is 
the  second  partial  product. 

Subtracting  this  second  partial  product  from  2x?y  +  3  xy2 
-t-y3,  we  have  xy^  +  y3  still  remaining  after  the  first  and  second 
partial  products  have  been  subtracted  from  the  whole  product. 
This  remainder  is  the  product  of  the  divisor  and  the  part  of  the 
quotient  not  yet  found.  Since  the  product  contains  the  term 
xy2  and  the  divisor  contains  the  term  x,  the  quotient  must  contain 
the  term  y2 ;  hence,  the  third  partial  product  is  xy2  +  y3. 

Subtracting  the  third  partial  product  the  remainder  is  zero. 
Hence  the  sum  of  the  three  partial  products  thus  obtained  is 
equal  to  the  whole  product,  and  it  follows  that  x2  +  2  xy  -f-  y2  is 
the  required  quotient. 


218  QUOTIENTS  AND  SQUARE  ROOTS 

162.  Problems  in  division  may  be  checked  by  substituting 
any  convenient  values  for  the  letters.  For  example,  in  this 
case,  x  =  l,  y  =  l,  reduces  the  dividend  to  8,  the  divisor  to  2, 
and  the  quotient  to  4,  which  verifies  the  correctness  of  the 
result. 

Since  division  by  zero  is  impossible  (see  Advanced  Course), 
care  must  be  taken  not  to  select  such  values  for  the  letters 
as  will  reduce  the  divisor  to  zero. 

Illustrative  Example  2.      Divide  2xi-\-xz  —  7  a?  +  5  x  —  1  by 

Solution.  [divisor. 

x2  +  2x-l, 


Dividend  or  product:  2  x*+xi  —  7  x2+5  x— 

1st  product,  2x2(x2+2a;-l):     2x*+ix8-2x2 12  x2-  3  x  +  1, 

Dividend  minus  1st  product :  —  3  x8  —  5  x2+ 5  x—  1         [quotient. 

2d  product,  -3  x(x2  +  2x-l):        -3x8-6x2+3x 
Dividend  minus  1st  and  2d  products  :  +i2+2i-1 

3d  product,  1  •  (x2  +  2  x- 1):  x2+2x-l 

Dividend  minus  1st,  2d,  and  3d  products :  0 

Check.    Substitute  x  =  2  in  dividend,  divisor,  and  quotient. 

Illustrative  Example  3.  Divide  20  a?  -  8  + 18  a*  +  22  a  - 19  a3 
by  2u2-3a  +  4. 

Solution.  Arranging  dividend  and  divisor  according  to  the  de- 
scending powers  of  a,  we  have  Tdivisor. 
Dividend  or  product:               18a4-19a8  +  20a2+22q-8[2  a2-3  a  +  4, 

1st  product :  18a4-27a8+36a2 9a2+4a-2, 

Dividend  minus  1st  product :  8  a3— 16  «2  +  22  a— 8      [quotient. 

2d  product :  8a8-12a2  +  16a  " 

Dividend  minus  1st  and  2d  products :       —  4a2+  6  a  — 8 
3d  product :  -  4a2+   6a-8 

Dividend  minus  all  products :  0 

Check.     Substitute  a  =  1  in  dividend,  divisor,  and  quotient. 


DIVISION  BY  A   POLYNOMIAL  219 

163.  From  a  consideration  of  the  preceding  examples  the 
process  of  dividing  by  a  polynomial  is  described  as  follows : 

1.  Arrange  the  terms  of  dividend  and  divisor  according  to 
descending  (or  ascending)  powers  of  some  common  letter. 

2.  Divide  the  first  term  of  the  dividend  by  the  first  term 
of  the  divisor.     This  quotient  is  the  first  term  of  the  quotient. 

3.  Multiply  the  first  term  of  the  quotient  by  the  divisor 
and  subtract  the  product  from  the  dividend. 

4.  Divide  the  first  term  of  this  remainder  by  the  first  term 
of  the  divisor,  obtaining  the  second  term  of  the  quotient. 
Multiply  the  divisor  by  the  second  term  of  the  quotient  and 
subtract,  obtaining  a  second  remainder. 

5.  Continue  in  this  manner  until  the  last  remainder  is  zero,  or 
until  a  remainder  is  found  whose  first  term  does  not  contain 
as  a  factor  the  first  term  of  the  divisor.  In  case  no  remainder 
is  zero,  the  division  is  not  exact. 

EXERCISES 

Check  the  result  in  each  case,  being  careful  to  substitute 
such  numbers  for  the  letters  as  do  not  make  the  divisor  zero. 

Divide  the  following : 

1.  a2  -f-  2  ab  +  b2  by  a  +  b. 

2.  a2 -2 ab  +  b2  by  a-b. 

3.  a3-  3  a2b  +  3  ab2  -W  by  a-  b. 

4.  2  a^  -f  2  x*y  —  4:  x2  —  x  —  4=  xy  —  y  by  x  -\-  y. 

5.  x3  -f-  %y2  —  x?y  —  y3  by  x  —  y. 

6.  tf  +  lrf  +  x  —  6bycc  +  3. 

7.  a^  +  4  xP  +  x  —  6  by  x  —  1. 

8.  :B4-6ar,  +  2ar!-3:K  +  6by  x  —  1. 


220  QUOTIENTS  AND  SQUARE  ROOTS 

9.  x3  +  3  x?y  -f-  3  xy2  +  y3  by  x2  +  2  an/  +  y2. 

10.  a?  — 8  3?  + 75  by  a?  — 5. 

11.  2  a? +  19  a2b +  9  ab2  by  2a+b. 

12.  a;4  —  4  ar3?/  +  6  a%2  —  4  a$3  +  ?/4  by  x  —  y. 

13.  x4  +  4 arfy -f 6 ar2/  +  4a#J-j-?/4byar!  +  2ar?/  +  2/2. 

14.  a;4  -f-  a^y  +  a;?/3  +  y*  by  a;  +  ?/. 

15.  x*  +  a%2  +  y*  by  x*  —  xy  + 1/2. 

16.  x*  —  y4  by  x  —  y. 

17.  a3  +  63  +  c3  —  3  a&c  by  a  +  6  +  c. 

18.  2a4+ll33-26a-2  +  16a;-3by  a^+7»-3. 

19.  x5  +  5  afy  + 10  x'y2  + 10  a,-2?/3  +  5  xy*+  y5  by  x2 +  2  xy  +  y2. 

20.  a,-5  -  x*  -  27  ar3  +  10  x2  -  30  x  -  200  by  a;2  -  4  x  - 10. 

21.  3a^  —  4a#  +  8a;z  —  4y2  +  8y«  —  3z2  by  x  —  2y  +  3z. 

22.  9rV-4r¥  +  4rs£2-s¥by3rs-2rt  +  s*. 

23.  9  a2b2  + 16  x2  -  4  a2  -  36  Vx2  by  3  a&  +  6  bx -  2a  -  4  a;. 

24.  a^  +  a?2?/  +  a^z  —  a;?/2  —  ?/2z  —  yz2  by  x2  — ?/z. 

25.  a5  +  a4&  +  a3  -  a362  -  2  a&2  +  &3  by  a2  +  ab-  b2. 

26.  a3  -{-  2Z3  —  2s  +  3  xyz  by  a;  +  y  —  z. 

27.  a3  +  &3  +  3  ab  - 1  by  a  +  b  —  1. 

28.  6 x5*- 11  a;4* +  23 0^  +  13^ -3 a;* +  2  by  3o*  +  2. 

29.  a3*  -  3  a2kbk  +  3  a*62*  -  ft3*  by  a*  -  6*. 

30.  32 sia  -  9  s^P  +  12  sH2h  -  18  saP - 17  £46  by -«•-*». 

164.  Since  the  dividend  is  the  product  of  the  divisor  and  quo- 
tient, it  follows  that  if  one  factor  of  an  expression  is  given,  the 
other  factor  may  be  found  by  division. 


SQUARE  ROOTS   OF  POLYNOMIALS  221 

EXERCISES 

Obtain  the  factors  of  each  of  the  following  products,  one 
factor  being  given  in  each  case : 

Product  Given  factor 

1.  x^  —  y8.  x  —  y. 

2.  ar3  +  y*.  x  +  y. 

3.  a5—  b5.  a  —  b. 

4.  a5  +  b5.  a  +  b. 

5.  a6  +  &6.  a?  +  b2. 

6.  a9  +  69.  a3  +  63. 

7.  r3  —  s3.  r2  +  rs  -{-  s?. 

8.  r4-s4.  r3  +  r2s  +  rs2+s3. 

9.  r5  4-  s5.  r4  —  ^s  -f  rV  —  rs3  -|-  s4. 

10.  a3-12a2  +  27a  +  40.  a-5. 

11.  ar*- 5  a;4?/ +  11  arty2  a2  -  3  xy  +  2 y2. 
— 14  a:2?/3  —  51  xy*  +  54  f/5. 

12.  a;4  +  xPy2  +  y*.  aP  —  xy  +  y2. 

13.  a3  +  5a2-2a-24.  a2  +  7a  +  12. 

14.  a5-5a4&  +  10a362  a2-2a6  +  62. 

- 10  a2b3  +  5ab*-  bs. 

15.  x5  —  5  x?y2  —  5  x2^3  +  y5.  x2  —  3  xy  +  y2. 

SQUARE  ROOTS  OF  POLYNOMIALS 

165.   In  §§  87,  88,  we  found  certain  trinomials  which  were 
perfect  squares,  namely, 

a2  +  2a&  +  62  =  (a  +  &)2,  (1) 

a2-2ab  +  b2=(a-b)2.  (2) 

Hence  we  know  the  square  roots  of  all  trinomials  which  are 
in  either  of  these  forms.     These  trinomial  squares  may  be  used 


222  QUOTIENTS  AND   SQUARE  BOOTS 

to  discover  a  process  for  finding  the  square  root  of  any  poly- 
nomial which  is  a  perfect  square. 

Finding  a  square  root  may  be  regarded  as  a  process  of 
division  in  which  divisor  and  quotient  are  equal  and  both  are 
to  be  found  simultaneously. 

Illustrative  Example.    Find  the  square  root  of  4arJ+12sc?/+9?/2. 

Considering  the  formula  (1)  we  are  to  pass  from  the  square 
a2  +  2  ab  +  b2  to  the  square  root  a  +  b,  and  for  this  purpose 
we  write  a2  +  2  ab  -f  b2  in  the  form  a2  +  6  (2  a  +  b)  and  arrange 
the  work  as  follows : 

Square  or  product,         4  x2  +  12  xy  +  9  y2\2  x  -f  3  y,  square  root. 
4  x2  1st  par'l  product. 

1st  par'l  divisor,        4  x 
1st  compl.  divisor,  4  a:  +  Sy 


\2  xy  +  9  y%  square  minus  1st  par'l  prod. 
12  xy  +  9  y  2,  2d  par'l  product. 


0 

Supposing  that  4  x2  is  the  a2  of  the  formula,  a  is  then  2  a, 
which  is  the  first  term  of  the  root.  Squaring  2  x  gives  4  x2,  the 
first  partial  product.  Subtracting  4  cc2  from  the  total  product 
leaves  12  #?/  +  9  y2,  which  is  the  b  (2  a  +  b)  of  the  formula. 

Since  b  is  not  yet  known,  we  cannot  find  completely  either 
of  the  factors  of  b  (2  a  +  b) ;  but  since  a  has  been  found,  we  can 
get  the  first  term  of  the  factor  2  a  -f-  b,  viz.  2aor2-2sc  =  4a;, 
which  is  the  first  partial  divisor.  Dividing  12  xy  by  4  a;  we 
have  3y,  which  is  the  b  of  the  formula.  Then  2a  +  b  =  4:X  +  3y 
the  first  complete  divisor. 

To  obtain  the  second  partial  product,  b  (2  a  +  b)  or  12  xy  +  9y2, 
we  multiply  4  x  +  3  ?/  by  3  y.  On  subtracting,  the  remainder  is 
zero  and  the  process  ends,  whence  the  required  root  is  2  x  +  3  y. 

It  should  be  clearly  understood  that  the  sum  of  the  first  and 
second  partial  products  is  a  square,  viz.,  the  square  of  2  x  +  3  y, 
because  it  has  been  constructed  just  as  a?  +  b(2  a  +  b)  was 
formed  from  a  +  &. 


SQUARE  ROOTS  OF  POLYNOMIALS  223 

EXERCISES 

Find  in  the  manner  just  described  the  square  roots  of  the 
following  trinomials : 

1.  36<c2-84a^  +  49?/2.  4.    81  m4+ 144  m?n2  +  64  n*. 

2.  16  a2  -  40  ab  +  25  b2.  5.   49  s6  -  84  s3  +  36. 

3.  121 t2  +  264  ta  +  144  a2.  6.    1  -  20  a4  + 100  a?. 

166.  The  process  just  applied  to  trinomials  is  applicable  to  poly- 
nomial squares  having  a  larger  number  of  terms,  by  regarding  the 
a  of  the  formula  at  each  step  as  representing  the  part  of  the  root 
already  found  and  b  as  the  term  of  the  root  about  to  be  found. 

Illustrative  Example.  Find  the  square  root  of  x2  +  2  xy  +  y2 
+  6  sc+  6  y  +  9.     The  work  may  be  arranged  as  follows : 

Square  or  product,  x2  +  2  xy  +  y2+Gx  +  Qy  +  9\x  +  y  +  3,         square  root. 
x2  1st  par'l  product. 


2xy  +  y2+6x+Qy  +  9,  1st  remainder. 

2xy  +  y2,  2d  par'l  product. 


6  x  +  6  y  +  9,  2d  remainder. 

6  x+  6  y+  9,  3d  par'l  product. 


1st  par'l  div'r,  2x 

1st  compl.  div'r,      2x  +  y 
2d  par'l  div'r,  2  x+  2  y 

2d  compl.  div'r,       2x  +  2y  +  3 

~ 0" 

After  the  first  two  terms  of  the  root  have  been  found,  namely, 
x  and  y,  then  we  consider  x  +  y  as  the  a  of  the  formula  and 
call  it  a'.  The  new  b,  which  we  call  b',  is  then  found  to  be  3. 
Subtracting  the  first  and  second  partial  products  is  the  same 
as  subtracting  (x-\-y)2,  that  is,  the  square  of  a'.  Hence,  the 
second  partial  divisor,  which  is  twice  a',  is  2  (x-\-y). 

In  case  there  are  four  terms  in  the  root,  then  the  sum  of  the  first 
three,  when  found  as  above,  is  regarded  as  the  new  a,  called  a". 
The  remaining  term  of  the  root  is  the  new  b,  and  is  called  b". 

The  formula  (a  —  b)2  =  a2  —  2  ab  +  62  is  not  needed,  since  this  may 
be  written  (a  +  (-  ft))2  =  a2  +  2  a  (-  b)  +  (-  by,  which  is  in  the 
form  of  (a  +  b)2  =  a2  +  2  ab  +  62. 


224  QUOTIENTS  AND  SQUARE  ROOTS 

EXERCISES 

Find  the  square  roots  of  the  following  polynomials: 

1.  x2  +  y2  +  z2-2xy  +  2xz-2yz. 

2.  9arJ  +  4?/2  +  z2  —  12xy  +  6xz  —  Ayz. 

3.  a?-8ab  +  16b2-2ac  +  c2  +  8bc. 

4.  a4  +  4a3  +  6a2  +  4a  +  l. 

5.  c4_4c3  +  6c2-4c  +  l. 

6.  ?/4  -  4  ?/2- 8  a-?/2 +  16  a +  163? +4. 

7.  x4  —  2  a8 +  3  a2  — 4*  +  4. 

8.  a2  +  a262  -  2  a26  +  2  ode  -  2  a&2c  +  &2c2. 

9.  a4  +  4a36  +  6a2&2  +  4a&3  +  &4. 

10.  a;4 — 4  a??/  +  6  x^y2 — 4  xy3  -f  2/4. 

11.  a;6  -  4  as"  +  4  x4  +  6  a;3  - 12  cc2  +  9. 

12.  a4  +  53  a2  +  14  a3  +  28  a  +  4. 

13.  a-2  +  16afy2  +  289  +  8a:2?/  +  34a:  +  136a:?/. 

14.  9  a4 +  4  a2 +  256 -12  a3 -96  a2 +  64  a. 

15.  a6  +  6a5  +  15a4  +  20a3  +  15a2  +  6a  +  l. 

16.  a6-6a5  +  15a4-20a3  +  15a2-6a  +  l. 

17.  16x6+4,y2-[-l-16xiy  +  8xi-4:y. 

18.  25 +  49  a2 +  4  a4 -70  a -20  a2 +  28  or5. 

19.  64  x6  +  192  x5  +  240  a4  +  160  ar5  +  60  x2  + 12  x  + 1. 

20.  4a6-12a,-5  +  13a;4-14a*!  +  13av!-4a;  +  4. 

21.  «4J4-2aV  +  a2-2fl26+2a6+52. 

22.  16a6  +  24a5  +  25a4  +  20a3  +  10a2  +  4a  +  l. 

23.  xHf  +  2 ary  +  3 a;4?/4  +  4 ary  +  3 tfy2 -\-2xy-\-l. 

24.  l  +  2a;  +  3ar2  +  4ar3  +  5ar4  +  4ar5  +  3a;6  +  2a;7  +  ar8. 

25.  9  a2 -  Gab  +  30 ac  +  6 ad  +  62- 10 6c -2  M  +  25C2 

+  10cd  +  d2. 


SQUARE  ROOTS   OF  ARABIC  NUMBERS  225 

SQUARE   ROOTS   OF   NUMBERS   EXPRESSED   IN   ARABIC   FIGURES 

167.  The  square  root  of  a  number  expressed  in  Arabic 
figures  may  be  found  by  the  process  just  used  for  polynomials. 

Illustrative  Example.     Find  the  square  root  of  5329. 

In  order  to  decide  how  many  digits  there  are  in  the  root  we 
observe  that  102  =  1000  and  1002  =  10000 ;  hence  the  root  lies 
between  10  and  100,  i.e.  it  contains  two  digits.  Since  802  = 
6400  and  702  =  4900,  we  see  that  7  is  the  largest  number  pos- 
sible in  tens'  place  of  the  root.  The  work  is  arranged  as 
follows : 


The  given  square, 

5329  |70  +  3,  square  root. 

a2  =  702 

4900,      1st  partial  product. 

2  a  =  2  x  70  =  140 

ft  =  3 

2  a  +  b  =  143 

429,      1st  remainder. 

429  =  6  (2  a  +  b). 

0 

Having  decided  as  above  that  the  a  of  the  formula  is  7  tens, 
we  square  this  and  subtract,  obtaining  429  as  the  remaining 
part  of  the  power. 

The  first  partial  divisor,  2  a  =  140,  is  divided  into  429  giving 
a  quotient  3,  which  is  the  b  of  the  formula.  Hence  the  first 
complete  divisor,  2  a  +  b,  is  143,  and  the  second  partial  prod- 
uct, b  (2  a  +  b),  is  429.  Since  the  remainder  is  zero,  the  pro- 
cess is  exact  and  73  is  the  square  root  sought. 

EXERCISES 

Find  the  square  roots  of  the  following  : 

1.  3249.       5.  6241.  9.  1849. 

2.  8836.       6.  7056.  10.  7225. 

3.  7569.       7.  9409  11.  3025. 

4.  8281.       8.  9801.  12.  9216. 


226  QUOTIENTS  AND  SQUARE  ROOTS 

168.  The  square  of  any  integer  from  1  to  9  contains  one  or 
two  digits ;  the  square  of  any  integer  from  10  to  99  contains 
three  or  four  digits ;  the  square  of  any  integer  from  100  to  999 
contains  five  or  six  digits ;  etc. 

Hence  it  is  evident  that,  if  the  digits  of  a  number  which  is 
a  perfect  square  be  separated  into  groups  of  two  each,  count- 
ing from  units'  place  toward  the  left,  the  number  of  groups 
thus  formed  is  the  same  as  the  number  of  digits  in  the  square 
root. 

Illustrative  Example.     Find  the  square  root  of  120,409. 

Since  this  number  is  divided  into  three  groups,  the  first  digit  is  in 
hundreds'  place.     The  work  is  arranged  as  follows : 

Square,  12  04  09  1300  +  40  +  7  =  347,  square  root. 


a2  =  3002        9  00  00,         1st  partial  product. 

2a  =  2  x  300=600 

b=   40 

2  a  +  b  =  640 

2  a'  =  2  x  340  =  680 


3  04  09,        1st  remainder. 
2  56  00  =  b  (2  a  +  b). 


V=     7 

2  a'  +  V  =  687 


48  09,       2d  remainder. 
48  09  =  V  (2  a'  +b'). 


0 

The  first  partial  divisor,  2  x  300,  is  completed  by  adding  the  second 
term  of  the  root,  40,  that  is,  2  a  +  b  —  600  +  40.  At  first  glance  it 
might  be  supposed  that  the  second  digit  is  5  instead  of  4,  since  600  is 
contained  in  30,409  50  times,  but  account  must  be  taken  of  the  addi- 
tion to  be  made  to  the  partial  divisor,  and  when  this  is  done  the 
quotient  is  40,  not  50. 

In  the  second  partial  divisor,  2  a'  stands  for  2  times  (300  +  40)  =  680, 
and  V  stands  for  the  third  digit  of  the  root. 

In  case  a  square  contains  an  odd  number  of  digits  the  last 
group  at  the  left  will  have  one  instead  of  two  digits. 

E.g.  3  47  21  has  three  places  in  its  square  root,  of  which  the  first, 
hundreds'  digit,  is  the  largest  square  in  3,  namely,  1. 


SQUARE  BOOTS   OF  ARABIC  NUMBEBS 


227 


EXERCISES 

Find  the  square  root  of  each  of  the  following : 

1.  294,849.  5.   3481.  9.    100,489.         13.   35,721. 

2.  37,636.  6.    7569.  10.     26,569.         14.   16,641. 

3.  872,356.  7.   1849.  11.   874,225.         15.   32,761. 

4.  599,076.  8.   73,441.        12.   170,569.         16.   223,729. 

169.  Since  the  square  of  any  decimal  fraction  has  twice  as 
many  places  as  the  given  decimal,  it  is  evident  that  the  square 
root  of  a  decimal  fraction  contains  one  decimal  place  for  every 
two  in  the  square. 

E.g.  (.15)2  =  .0225 ;  (.012)2  =  .000144. 

Hence,  for  the  purpose  of  determining  the  decimal  places  in 
the  root,  the  decimal  part  of  a  square  should  be  divided  into 
groups  of  two  digits  each,  counting  from  the  decimal  point 
toward  the  right. 

Illustrative  Example.  Find  the  square  root  of  4.6225.  Ac- 
cording to  §§  168, 169  the  root  contains  one  digit  in  the  integral 
part  and  two  in  the  decimal  part.     The  work  is  as  follows : 


a2  =  22 
2a  =  2  x  2 
b  = 
2a  +  b  = 


4.62  25 1  2  +  .1  +  .05  =  2.15 
4 


4.0 
4.1 


2  a'  =  2  x  2.1  =  4.2 
b'=  _J;05 

2  a'  +  b'=  4.25 


.62  25,     first  remainder. 


.41 =  b  (2  a  +  b). 


.21  25,     second  remainder. 
.21  25  =  b'  (2  a'  +  b'). 


0 


This  process  is  also  applicable  for  the  purpose  of  approxi- 
mating the  square  root  of  a  number  which  is  not  a  perfect 
square 


228 


QUOTIENTS  AND   SQUARE  ROOTS 


Illustrative  Example.  Find  the  approximate  square  root  of 
582  to  three  decimal  places.  The  solution  below  shows  all  the 
steps  of  the  work. 

5  82  1 20  +  4  +  .1  +  .02  +  .004  =  24.124 
4  00 


a2  =  202 

2  a  =  2  x  20 

=  40. 

b  = 
2a  +  b 

'      4. 
=  44. 

2  a'  =  2  x  24 

=  48. 

b'  = 

.1 

2  a'  +  V  = 

48.1 

2  a"  =  2  x  24.1 

=  48.2 

b"  = 

.02 

2  a"  +  b"  = 

48.22 

182, 
176 


6.00, 


first  remainder. 

=  b  (2  a  +  b). 

second  remainder. 


4.81         =b'  (2  a'  +  6'). 
1.1900,  third  remainder. 


.9644      =  b"  (2  a"  +  b"). 


2  a"'  =  2  x  24.12  =  48.24 

V"  =  .004 

2  a'"  +  V"  =  48.244 


.225600,        fourth  remainder. 


.192976  =  V"  (2  a'"  +  6'"). 


.032624 


The  decimal  points  are  handled  exactly  as  in  division  of  deci- 
mals in  arithmetic,  the  chief  care  being  needed  in  forming  the 
divisors. 

170.  Evidently  the  process  in  this  example  may  be  carried 
on  indefinitely.  24.124  is  an  approximation  to  the  square  root 
of  582.  In  fact,  the  square  of  24.124  differs  from  582  by  the 
small  fraction  .032624.  24.12  is  the  nearest  approximation 
using  two  decimal  places.  If  the  third  figure  were  5  or  any 
digit  greater  than  5,  then  24.13  would  be  the  nearest  approxi- 
mation using  two  decimal  places.  Hence  three  places  must  be 
found  in  order  to  be  sure  of  the  nearest  approximation  to  two 
places. 


SIMPLIFYING   RADICALS  229 


EXERCISES 

Find  the  square 

roots  of  the  folio 

wing, 

correct  to  two  deci 

mal  places : 

1.    387. 

7.    2. 

13.    .02. 

2.    5276. 

8.  3. 

14.    .003. 

3.   2.92. 

9.    5. 

15.   .5. 

4.   27.29. 

10.   7. 

16.    .005. 

5.   51. 

11.   8. 

17.    .307. 

6.   3.824. 

12.    11. 

18.  200.002. 

SQUARE  ROOTS  OF  FRACTIONS  EXPRESSED   IN  ARABIC  FIGURES 

171.  Since  in  arithmetic  the  product  of  two  fractions  is 
found  by  multiplying  their  numerators  and  their  denominators, 
a  fraction  is  squared  by  squaring  its  numerator  and  its  denom- 
inator separately. 

Hence,  to  extract  the  square  root  of  a  fraction,  we  find  the 
square  root  of  its  numerator  and  its  denominator  separately. 

E.g.  Vif  =  |,  since  f  x  |  =  if. 

However,  in  approximating  the  square  root  of  a  fraction  whose 
denominator  is  not  a  perfect  square,  if  the  final  result  is  to  be 
reduced  to  a  decimal,  the  direct  use  of  this  method  is  cumber- 
some since  it  usually  involves  the  approximation  of  two  square 
roots  and  always  necessitates  dividing  by  a  long  decimal  fraction. 

E.g.  Vf  =  a/2  ■+■  V3  =  1.4142  +  1.7321,  in  which  we  are  now 
obliged  to  divide  by  1.7321. 

This  difficulty  may  be  avoided  in  either  of  two  ways : 

1.  The  fraction  may  be  reduced  to  a  decimal  before  the 
root  is  approximated. 


E.g.  Vy  =  V.666  •  •  •  =  .8165. 


230  QUOTIENTS  AND   SQUARE  BOOTS 

2.  The  denominator  of  the  fraction  may  be  made  a  perfect 
square  before  approximating  the  root. 

y     *3      *9        3  3 

By  either  of  these  two  processes  only  one  square  root  is 
approximated,  and  there  is  only  a  short  division  by  3  instead 
of  a  long  division  by  1.7321. 

It  is  clear  that  any  fraction  can  be  changed  into  an  equal  fraction 
whose  denominator  is  a  perfect  square  by  multiplying  numerator  and 
denominator  by  the  proper  number. 

E-ff-     J  =  h  I  =  M>  j  =  il>  etc.     If  a  fraction  is  given  in  the  form 

— ,  it  may  be  written  Vf  =  vf  =  |V3.      In  like  manner, 
V3 

J__  =  7.  _J_  =  7  JT=7  JH=7  ^ 

vn       vn      vn      Xi-Ji    ii 


EXERCISES 

Find  approximately  correct  to  two  decimal  places  the  follow- 
ing square  roots. 

In  the  first  ten  obtain  the  results  in  three  different  ways : 
(a)  Find  the  root  of  each  numerator  and  denominator  separately ; 
(6)  reduce  each  fraction  to  a  decimal ;  (c)  reduce  each  fraction 
so  as  to  make  its  denominator  a  perfect  square. 

•In  the  remaining  exercises  use  methods  (6)  and  (c)  only. 
In  each  case  compare  the  results  obtained. 


5. 


v|. 

6.   VJ. 

ii.  V3?. 

16. 

Vf. 

Vf. 

Vf 

7.   V*f. 

8.  vjf 

12.  V^. 

13.  V|. 

17. 

1 

V5* 

9.   V|. 

io.  Vf 

14.  Vf. 

15.  Vf 

18. 

5 
V13 

19. 


3.   V*.  8.   V*f.        13.    Vf.  V5  20. 


21. 


3 

V7* 

7 
y/Tf' 

Vf 


SIMPLIFYING  RADICALS  231 

172.  Principle  XVIII  may  be  used  to  advantage  in  approxi- 
mating the  square  roots  of  certain  integral  numbers. 

E.g.  suppose  V2  has  been  computed,  and  VS  is  desired.  It  is 
unnecessary  to  compute  the  V8  directly,  for  by  XVIII, 

VS=  vTT2  =  y/i-  V2  =  2V2. 

This  sort  of  simplification  is  possible  whenever  the  number 
under  the  radical  sign  can  be  resolved  into  two  factors,  one  of 
which  is  a  perfect  square. 

E.g.   suppose  the  V5  to  have  been  computed,  then 


VT25  a  V25V5  =  V25  •  V5  =  5  V5. 

In  like  manner,  Va568  may  be  written 

Va462  •  ab  =  Va*P  ■  Vab  =  a?bVab. 

Definition.  A  radical  expression  is  said  to  be  simplified 
when  the  number  under  the  radical  sign  is  in  the  integral  form 
and  contains  no  factor  which  is  a  perfect  square. 

E.g.    the  simplified  forms  of  Vl^5>  VaW  VT  — ~> are  respectively, 
5  V5,  a%  Vab,  i  V5,  i  >/3. 

EXERCISES 

Given  V2  =  1.4142,  V3  =  1.7321,  V5  =  2.2361,  compute 
the  following,  correct  to  three  places  of  decimals,  without 
further  extraction  of  roots : 


1. 

V80. 

6. 

V2-3. 

11. 

V27  +  V|. 

2. 

vj. 

7. 

V72. 

12. 

V45  +  VJ. 

3. 

VJ- 

8. 

V98. 

13. 

V50-VJ  +  V8. 

4. 

V48. 

9. 

V363. 

14. 

V48+V75-V3. 

5. 

V75. 

10. 

VI25. 

15. 

V32+V72-V18. 

232  QUOTIENTS  AND   SQUARE  ROOTS 

Simplify  the  following : 


16.  V32a26.  19.  VioVy^.        22.  V500  x7  asb. 

17.  VSlxW2.  20.  V63  b(fd\         23.   -yJ'Stf  +  Kxy  +  dy2. 

18.  VoO  aW.  21.  V900  o6V.       24.   V8«2-12y2. 

25.  V32a2-64a6  +  32  62.  26.  VF25  x2  +  250  ajy  + 125  y\ 

27.  Find  approximately  to  four  decimal  places  the  sides  of 
a  square  whose  area  is  120. 

28.  Approximate  to  four  decimals  the  side  of  a  square  having 
an  area  equal  to  that  of  a  rectangle  whose  sides  are  15  and  20. 

29.  How  many  rods  of  fence  are  required  to  fence  a  square 
piece  of  land  containing  50  acres,  each  acre  containing  160 
square  rods? 

30.  A  square  checkerboard  has  an  area  of  324  square  inches. 
What  are  its  dimensions  ? 

173.  In  adding  or  subtracting  expressions  containing  radi- 
cals it  is  always  best  to  first  reduce  each  radical  expression  to 
its  simplest  form,  since  this  often  gives  opportunity  to  com- 
bine terms  which  are  similar  with  respect  to  some  radical 
expression. 

Ex.1.  V32+ V72- Vl8  =  4V2  +  6V2-3V2  =  7V2by 
Principles  XIII,  I,  and  II. 

Ex.2.    VJ+Vi2-V|  =  |V3  +  2v/3-|V3 

=  a  +  2-i)V3=l|V3. 

EXERCISES 

Simplify  each  of  the  following  as  far  as  possible  without 
approximating  roots. 

1.  V27  +  2V48-3V75. 

2.  V20  +  V125-  VT80. 

3.  3V432-4V3  +  V147. 


APPLICATIONS   OF  SQUARE  BOOT  233 


4.  3V2450-25V2  +  4V13122 

5.  3y2^xJz  +  2^xJz?-yziyJ-2 

6.  V4  x*y  +  V2o  a-?/3  —  x  -\/xy 


7.  Vaa-2  —  bar  +  V4  cwV  —  4 

8.  4A/|-fV^-3V27. 

9.  2  V|  +  V60  +  Vf. 

10.  5  V3- 2  V48 +  7  VIM. 


11.  Va3  -  a2b  -  Vai*  -  b3  -  V(a  +  6)(a2  -  62). 

12.  A/a  +  3  V2~cT- 2 V3a  +  V4~a-  V8a+  Vl2a. 


13.    Va-3  -\-2xry  +  xy2  —  Va?  —  2  x2y  +  a;?/2  —  V4  xy\ 


14.    Vr  -  s  +  Vl6  r  -  16  s  +  Vrt2  -  st2  -  V9(r— *). 


15.    V(m  —  w)2a  +  V(m  +  n)2a  —  -\/am2  +  Va  (1  —  ra)2  — 


16.    V32  x2y*  +  V162  arY  -  V512  arfy4  +  V1250  tfy*. 

APPLICATIONS  OF   SQUARE  ROOT 

174.  Some  of  the  most  interesting  and  useful  applications  of 
the  square  root  process  are  concerned  with  the  sides  and  areas 
of  triangles. 

The  fact  that  the  sum  of  the  squares  on  the  two  sides  of  a 
right  triangle  equals  the  square  on  the  hypotenuse  was  used 
in  Chapter  VI.     (Pythagorean  Proposition,  page  206.) 

If  a  and  b  are  the  lengths  of  the  sides,  and  c  the  length  of 
the  hypotenuse,  all  measured  in  the  same  unit,  this  propo- 
sitionsays:  <?  =  a2  +  b\  (1) 

Hence,  by  S,  a2  =  <?  -  b\  (2) 

and  b2  =  c2  -  a2.  (3) 


234  QUOTIENTS  AND   SQUARE  EOOTS 

Taking   the   square   root   of   both    sides   in   each   of  these 

equations, 

c=Va2  +  62.  (4) 

a  =  Vc2^^62.  (5) 

b  =  V^^a2.  (6) 

The  negative  square  root  is  omitted  here,  as  a  negative 
length  cannot  apply  to  the  side  of  a  triangle.  By  these  formu- 
las, if  any  two  sides  of  a  right  triangle  are  given,  the  other 
may  be  found. 

E.g.   if  a  =  4,  b  =  3,  then,  by  (4), 

c  =  V42  +  32  =  Vl6  +  9  =  V25  =  5. 

If  c  =  5,  b  =  3,  then,  by  (5), 


a  =  V52  -  32  =  V25  -  9  =  Vl6  =  4. 

If  c  =  5,  a  =  4,  then,  by  (6), 

b  =  Vo2-42  =  V25  -  16  =  \/9  =  3. 

Illustrative  Problem.  If  the  two  sides  of  a  right  triangle  are 
8  and  12,  find  the  hypotenuse  correct  to  two  decimal  places. 

Solution.     We  have  c  -  Va2  +  b2  =  V64  +  144  =  V208, 

V208  =  VI6TT3  =  VIS  •  V13  =  4VI3  =  4(3.605)=  14.420. 

PROBLEMS 

In  solving  the  following  problems,  simplify  each  expression 
under  the  radical  sign  before  extracting  the  root.  Find  all 
results  correct  to  two  decimal  places.  In  each  case  construct 
a  figure. 

1.  The  sides  about  the  right  angle  of  a  right  triangle  are 
each  15  inches.     Find  the  hypotenuse. 

2.  The  hypotenuse  of  a  right  triangle  is  9  inches  and  one 
of  the  sides  is  6  inches.     Find  the  other  side. 


APPLICATIONS   OF  SQUARE  ROOT  235 

3.  The  hypotenuse  of  a  right  triangle  is  25  feet  and  one 
of  the  sides  is  15  feet.     Find  the  other  side. 

4.  The  hypotenuse  of  a  right  triangle  is  7  rods  and  one  of 
the  sides  is  5  rods.     Find  the  other  side. 

5.  The  hypotenuse  of  a  right  triangle  is  12  inches  and  the 
two  sides  are  equal.     Find  their  length. 

Let  s  equal  the  length  of  one  of  the  equal  sides. 

Then  s2+s2=144. 

2  s2  =  144. 

s2  =  72. 

s  =  V72  =  6V2=  6  x  1.414  =  8.484. 

6.  The  hypotenuse  of  a  right  triangle  is  30  feet  and  the 
sides  are  equal.     Find  their  length. 

7.  The  hypotenuse  of  a  right  triangle  is  h  and  the  sides 
are  equal.  Find  their  length.  Solve  5  and  6  by  means  of 
the  formula  here  obtained. 

8.  The  diagonal  of  a  square  is  8  feet.     Find  its  area. 

9.  The  diagonal  of  a  square  is  d.  Find  an  expression  in 
terms  of  d  representing  its  area. 

10.  The  side  of  an  equilateral  triangle  is  6  inches.     Find 
the  altitude. 

A   line   drawn  from   a  vertex   of   an  equi- 
lateral triangle  perpendicular  to  the  base  meets  / 
the    base    at    its    middle    point.      Hence    this             / 
problem  becomes  :   the  hypotenuse  of  a  right           / 
triangle   is   6   and   one   side   is  3.     Find  the        / 
remaining  side. 

11.  The  side  of  an  equilateral  triangle  is  10.    Find  the  alti- 
tude. 


236  QUOTIENTS  AND  SQUARE  BOOTS 

12.   The  side  of  an  equilateral  triangle  is  s.     Find  the 
altitude. 

This  is  equivalent  to  finding  a  side  of  a  right  triangle  whose  hypote- 
nuse is  s,  the  other  side   being  -.    Let  a  equal  altitude. 


Then  •  =  V"-(§)'-V^ 


;V4-^=V¥!=V^ 


-V 


j-VS=fv^. 


This  formula  gives  the  altitude  of  any  equilateral  triangle  in 
terms  of  the  side.  By  means  of  this  formula  solve  11  and  12. 
It  is  interesting  to  notice  that  the  square  root  of  3  is  the  only 
root  required  in  finding  the  altitude  of  any  equilateral  triangle 
whatever. 

13.  Find  the  altitude  of  an  equilateral  triangle  whose  side 
is  41.     Substitute  in  the  formula  under  12. 

14.  Find  the  area  of  an  equilateral  triangle  whose  side  is  5. 

Since  the  area  of  a  triangle  is  \  the  product  of  the  base  and  alti- 
tude, we  first  find  the  altitude  by  means  of  the  formula  under  12,  and 
then  multiply  by  J  the  base. 

15.  Find  the  area  of  the  equilateral  triangle  whose  side  is  s. 

s2    — 
Show  the  result  to  be  — V3. 
4 

16.  If  the  area  of  an  equilateral  triangle  is  16  square 
inches,  find  the  length  of  the  side. 

Let  s  equal  the  length  of  the  side.     Then  bv  the  formula  derived 

s2         

under  15,  16  =  —  V3. 

Hence  (§§  171,  172),  s2  =  -^4  =  —  V3  =  21.33  x  1.732. 
V3      3 


APPLICATIONS   OF  SQUABE  BOOT  237 

17.  The  area  of  an  equilateral  triangle  is  50  square  inches. 
Find  its  side  and  altitude. 

18.  The  area  of  an  equilateral  triangle  is  a  square  inches. 
Find  the  side. 

Solve  the  equation  a  =  —  V3  f or  s,  and  simplify  the  expression, 


finding    s2  =  -J ,  and  s  =  V  ~V~  =  -  V3aV3. 

V3  0  6 

19.  The  area  of  an  equilateral  triangle  is  240  square  inches. 
Find  its  side.     (Substitute  in  the  formula  obtained  under  18). 

20.  Find  the  area  of  a  regular  hexagon  whose 
side  is  7. 

A  regular  hexagon  is  composed  of  6  equal  equi- 
lateral triangles,  whose  sides  are  each  equal  to  the 
side  of  the  hexagon  (see  figure).  Hence  this  prob- 
lem may  be  solved  by  finding  the  area  of  an  equi- 
lateral triangle  whose  side  is  7,  and  multiplying  the  result  by  6. 

21.  Find  the  area  of  a  regular  hexagon  whose  side  is  s. 
(Solve  20  by  substituting  in  the  formula  obtained  here.) 

22.  The  area  of  a  regular  hexagon  is  108  square  inches. 
Find  its  side. 

If  the  area  of  the  hexagon  is  108  square  inches,  the  area  of  one  of 
the  equilateral  triangles  is  18  square  inches.  Hence  this  problem  can 
be  solved  like  18. 

23.  The  area  of  a  regular  hexagon  is  a  square  inches.  Find 
its  side.  (Solve  22  by  substituting  in  the  formula  obtained 
here.) 

24.  Find  the  radius  of  a  circle  whose  area  is  9  square  inches. 

The  area  of  a  circle  is  found  by  squaring  the  radius  and  multiply- 
ing by  3.1416.  The  number  3.1416  is  approximately  the  quotient 
obtained  by  dividing  the  length  of  the  circumference  by  the  diameter 
of  the  circle.     This  quotient  is  represented  by  the  Greek  letter  tt 


238 


QUOTIENTS  AND   SQUARE  BOOTS 


(pronounced  pi).  In  this  chapter  we  use  3^  as  an  approximation  to  ir. 
This  differs  from  the  real  value  of  it  by  less  than  .0013,  and  hence  is 
accurate  enough  for  most  purposes.  If  a  represents  the  area  of  a 
circle,  the  above  rule  may  be  written 


a 


irr'. 


Hence  if  a  =  9, 
and 


9  =  9^  _  63_  _  2  g63 

j     3f     22  ' 


25.  Find  the  radius  of  a  circle  whose  area  is  68  square  feet. 

26.  Find  the  radius  of  a  circle  whose  area  is  a  square  feet. 


We  have 


Hence 


a  =  irri,  or  rz  = 


1  7T  \  7H  IT 


In  problems  stated  in  terms  of  letters,  the  results,  of  course,  cannot 
be  reduced  to  a  decimal.  In  such  formulas  it  is  best  not  to  replace 
the  letter  ir  by  any  of  its  approximations. 

27.  Find  the  sum  of  the  areas  of  a  circle 
of  radius  6  and  the  square  circumscribed 
about  the  circle. 

The  area  of  the  circle  is  627r  =  36  ir,  and  the 
area  of  the  square  is  4  •  62  =  4  •  36  ;  i.e.  the 
square  contains  4  squares  whose  sides  are  6. 
The  sum  of  the  areas  is 

4  •  36  +  36  7T  =  (4  +  tt)  36  =  (4  +  3f)  36. 

28.  Find  an  expression  for  the  sum  of  the  areas  of  a  circle 
of  radius  r  and  the  circumscribed  square.  (Solve  27  by  sub- 
stituting in  the  formula  here  obtained.) 

29.  If  the  sum  of  the  areas  of  a  circle  and  the  circumscribed 
square  is  64,  find  the  radius  of  the  circle. 

By  the  formula  obtained  under  28, 

64=  (4  +  7r)r2  =  ^r2. 


Hence, 


r2  =  64^  =  8.96, 
r  =  V8.96  =  2.99. 


APPLICATIONS  OF  SQUARE  ROOT 


239 


30.  If  the  sum  of  the  areas  of  a  circle  and  the  circumscribed 
square  is  640  square  feet,  find  the  radius  of  the  circle. 

31.  The  sum  of  the  areas  of  a  circle  and  the  circum- 
scribed square  is  a.  Find  an  expression  representing  the 
radius  of  the  circle.     (Replace  w  by  3|  before  simplifying.) 

32.  If  the  radius  of  a  circle  is  12,  find 
the  difference  between  the  area  of  a  circle 
and  the  circumscribed  square. 

33.  If  the  radius  of  a  circle  is  r,  find  the 
difference  between  the  area  of  the  circle  and 
the  circumscribed  square.  (Solve  32  by 
substituting  in  the  formula  obtained  here.) 

34.  If  the  radius  of  a  circle  is  16,  find  the 

area  of  the  inscribed  square.    (This  is  the  same  problem  as  finding 
the  area  of  a  square  whose  diagonal  is  32.    See  problems  8  and  9.) 

35.  If  the  radius  of  a  circle  is  r,  find  an  expression  repre- 
senting the  area  of  the  inscribed  square. 

(This  is  problem  9,  the  hypotenuse  being 
2r.) 

36.  If  the  radius  of  a  circle  is  12,  find 
the  difference  between  the  area  of  the  circle 
and  the  area  of  the  inscribed  square. 

37.  If  the  radius  of  a  circle  is  r,  find 
an  expression  representing  the  differences 

between  the  areas  of  the  circle  and  the  inscribed  square. 

38.  The  radius  of  a  circle  is  10.  Find  the  area  of  an 
inscribed  hexagon.    See  note,  problem  20. 

39.  The  radius  of  a  circle  is  6.  Find 
the  difference  between  the  areas  of  the 
circle  and  the  inscribed  hexagon. 

40.  Find  an  expression  representing  the 
difference  between  the  areas  of  a  circle  with 
radius  r  and  the  inscribed  regular  hexagon. 


240  QUOTIENTS  AND  SQUARE  ROOTS 

SOLUTION  OF  QUADRATIC  EQUATIONS  BY  MEANS   OF  SQUARE  ROOT 

175.  Illustrative  Problem.  A  rectangular  field  is  18  rods 
longer  than  it  is  wide  and  its  area  is  50  acres.  What  are  its 
dimensions  ? 

Let  x  =  the  width  of  the  field. 

Then,  x  +  18  =  the  length  of  the  field. 

And  x  (x  +  18)  =  the  number  of  square  rods  in  the  field. 

Hence,  rr2  +  18  x  =  8000,                               (1) 

or  x2+  18  x-  8000  =  0.                                   (2) 

We  are  not  able  to  factor  the  left  member  of  the  equation 
by  any  method  thus  far  known,  and  hence  we  cannot  solve  the 
equation  as  in  §  145.  We  therefore  proceed  to  study  a  general 
method  of  solving  quadratic  equations. 

Consider  the  equation  in  the  form  (1)  above.  We  seek  a 
number  k2  such  that  x2  + 18  x  +  k2  shall  be  a  perfect  square. 
By  §  134  the  middle  term  of  a  trinomial  square  is  twice  the 
product  of  the  square  roots  of  the  two  square  terms.  Hence, 
18  x=2  kx,  that  is,  k  must  equal  9. 

Hence,  adding  92(=k2)  to  each  member  of  (1), 

a^+18.T  +  92  =  8000  +  92,  (3) 

or,  x2  +  18  x  +  81  =  8081.  (4) 

Since  the  left  member  is  now  a  perfect  square,  we  may 
extract  the  square  root  of  both  sides,  approximating  the  root 
on  the  right. 


Hence,  x  +  9  =  ±  V8081  =  ±  89.89, 

giving  x  =  -  9  +  89.89  =  80. 89, 

and  also,  x  =  -  9  -  89.89  =  -  98.98. 

In  this  case  the  negative  result  is  not  applicable  to  the 
problem.  Hence  the  width  of  the  field  is  80.89  rods,  which 
is  correct  to  two  decimal  places. 


QUADRATIC  EQUATIONS  241 

Illustrative  Example.     Solve  the  equation : 

x2  -  12  x  +  42  =  56.  (1) 

By  S,  x2-12x  =  14.                                                  (2) 

By  A ,  x2  -  12  x  +  k2  =  14  +  F.                                         (3) 

By  §  175,  -  12z  =  2  fee  or  *=  -  6. 

Hence,  x2  -  12  x  +(-  6)2  =  14  +  36  =  50.                                (4) 

Taking  square  roots,       x  —  6  =  ±  V50  =  ±  5  V2.  (5) 

By  ,4,  *  as  6±  7.071.                                      (6) 

Hence  x  =  6  +  7.071  =  13.071, 

and  also  x  =  6  -  7.071  =  -  1.071. 

176.  This  process  is  called  solving  the  quadratic  equation  by 
completing  the  square,  since  in  each  case  a  number  is  added  to 
both  sides  which  makes  the  left  member  a  trinomial  square. 

Since  the  process  always  involves  extracting  the  square  root 
in  order  to  find  the  value  of  the  unknown,  the  two  solutions 
of  a  quadratic  equation  are  commonly  called  the  roots  of  the 
equation.  By  analogy  the  solutions  of  any  equation  are  some- 
times called  its  roots. 

EXERCISES 

In  solving  the  following  quadratic  equations  the  result  may 
in  each  case  be  reduced  so  that  the  number  remaining  under 
the  radical  sign  shall  be  2,  3,  or  5.  (§  172.)  Use  these 
square  roots  only. 

1.  x2-4a;  =  8.  9.  a2- 4a;  =  16. 

2.  JB»=3-6aJ.  10.  2x  =14  +  4  x. 

3.  4 a;  =  16 -a8.  11.  24  =  3 x2  + 12 x. 

4.  a;2 +  6  a;  =  9.  12.  69  -  18  x  =  3  a;2. 

5.  a;2  +  6a;  =  ll.  13.  84  +  24  x  =  12  x>. 

6.  af-^x  =  12.  14.  25-3^  =  5*;. 

7.  a?-8a;=-14.  15.  x2  +  ix  =  2. 

8.  aj8  =  2a;+l.  16.  a^-fa;  =  -2^. 


242  QUOTIENTS  AND   SQUARE  ROOTS 

177.  In  case  the  coefficient  of  x2  is  not  unity,  both  members 
may  be  divided  by  this  coefficient. 

Example.    Solve  3  x2  +  8  x  =  4.  (1) 

By  A  i»+J*  =  f  (2) 

By  A,  *P  +  |* +*»  =  !  +  *■.  (3) 

By  §  175,  §x  =  2£xor&=f. 

Hence,  **+  t*  +  (*)?  =  ♦  +  V  =  ¥•  (4) 

Taking  square  roots,  *  +  f  =  ±  V-^-  (5) 

Hence,  x=  -|±|V7^  (6) 

and  the  two  roots  are  x  =  0.43, 

and  x  =  —  3.10. 

The  preceding  example  may  also  be  solved  as  follows  : 

Multiplying  each  member  of  (1)  by  4  •  3  =  12, 
then,  36  x2  +  96  x  =  48. 

By  A,  36  x2  +  96  x  +  t*  =  48  +  k2, 

where  -  12  kx  =  96  x  or  &  =  8. 

Hence,      36  xa  +  96  x  +  (8)2  =  48  +  64  =  112, 
and  6  x  +  8  =  ±  VTl2  =  ±  4  VT. 

Therefore  x  =  -  f  ±  ]  VT. 

178.  The  advantage  of  this  solution  is  that  fractions  are 
avoided  until  the  last  step,  and  the  value  of  k  is  found  to  be 
the  same  as  the  coefficient  of  x  in  the  given  equation.  This 
may  always  be  accomplished  by  multiplying  the  members  of 
the  given  equation  by  four  times  the  coefficient  ofx2. 

In  the  solution  of  the  following  equations  only  the  square 
roots  V2,  V3,  V5,  V6,  V7,  need  be  used.  In  all  cases  where 
the  roots  are  integers  or  exact  fractions  the  solution  may  also 
be  obtained  by  factoring  as  in  §  145. 

1.  2x*  +  Zx=2.         4.   6x  +  l  =  -Sx2.     7.   2x2-3x  =  U. 

2.  Sx2  +  5x  =  2.         5.   2ar2  =  5z  +  3.         8.   3^  =  9  +  2^. 

3.  3z  =  9-2ar!.  6.   ±x  =  2x2-l.         9.   4x2  =  2x  +  l- 


QUADRATIC  EQUATIONS  243 

10.  §x-l  =  3x2.       23.   2z-l  =  -4 a2.  36.  6^-12^  =  2. 

11.  2^  +  4^  =  23.     24.   5x»+16:r=-2.  37.  3x2  +  2x  =  5. 

12.  3^-7  =  40;.        25.   4a*  +  l  =  8ar.  38.  2  +  3x  =  2x>. 

13.  2x*-5  =  3x.       26.    2^-3^  =  20.  39.  8«  +  l  =  -4arJ. 

14.  4ar>  =  6a;-l.        27.   2arJ-3  =  -5z.  40.  8  +  4a;  =  3a;2. 

15.  2x  =  l-5x2.       28.   3ar  +  4a;  =  8.  41.  10  +  4^  =  5^. 

16.  3*-20«-2a£   29.    10-4:r  =  5a;*.  42.  2  +  5^  =  3^. 

17.  2x  +  3x2  =  9.       30.    l+4x2  =  -6x.  43.  3x-l±  =  2x2. 

18.  4a^-l  =  3ic.        31.   5-3a  =  2z2.  44.  3x2-2x  =  5. 

19.  4z  =  7-2;r!.       32.    7  +  4x  =  2xi.  45.  2a^+4*»l. 

20.  2x+l  =  bx2.        33.    6  a2 +  12  a  =  2.  46.  3x-l  =  -4ors. 

21.  3x2  +  4x=7.       34.   6x2-12x=-2.  47.  23  +  4 a  =  2 a2. 

22.  3z  +  9  =  2x2.       35.    &x2+12x=-2.  48.  Zx-l  =  2x2. 

179.   Solution  by  Formula.    Solve  the  equation 

ax2  +  bx  +  c  =  0.  (1) 


By  S,  M, 

4  a2x2  +  4  aftx  =  —  4  ac. 

(2) 

By  ,4, 

4  a2x2  +  4  aftx  +  £2  =  -  4  ac  +  k2, 

(3) 

where 

4  a£x  B  4  aix  or  k  =  b. 

Hence 

4  a2x2  +  4  a&x  +  b2  =  b2  -  4  ac. 

.(4) 

Taking  square  roots,          2  ax  +  b  =  ±  \/&2  —  4  crc. 

(8) 

By  5,  A 

_  -  6  ±  Vb2  -  4  ac 

(6) 

Calling  the  two  values  of  x  in  the  result  xx  and  x2  we  have, 


_-6+V62-4ac        _-6-V62-4gC 
ri~~     ~2«r~    ~;  Jr2~~    ~2^~     "7 


244  QUOTIENTS  AND   SQUARE  ROOTS 

Any  quadratic  equation  may  be  reduced  to  the  form  of  (1) 
by  simplifying  and  collecting  the  coefficients  of  x2  and  x.  Hence 
any  quadratic  equation  may  be  solved  by  substituting  in  the 
formulas  just  obtained. 

Ex.  1.     Solve  x2-4x  +  l  =  0. 

In  this  case  a  =  1,  b  =  —4,  c  =  1. 


Hence 

,    _-(-4)±V(-4)2_4.1.1 

2-1 
x1  =  2  +  VS 

From  which 

and 

x2  =  2  -  V3. 

Ex.  2.     Solve 

3a:2  +  16  x  -  12  =  0. 

Here 

a  =  3,  b  =  16,  c  =  -  12. 

Here 

_  _  16  ±  V(16)2-4-3(-  12) 

2-3 

From  which 

Zj  =  §,  and  x2  —  —  6. 

180.  A  quadratic  equation  may  be  proposed  for  solution 
which  has  no  roots  expressible  in  terms  of  the  numbers  of 
arithmetic  or  algebra  so  far  as  yet  studied. 

Example.     Solve  x2  +  4  x  =  -  8.  (1) 

By  .4,  a:2  +  4x  +  4=-4.  (2) 

Taking  square  roots,  x  +  2  =  ±  V—  4.  (3) 

V—  4  is  unknown  to  us  as  a  number  symbol,  since  there  is  no  num- 
ber thus  far  considered  whose  square  equals  —  4.  (See  Principle  XI.) 
Such  symbols  are  defined  and  used  in  the  Advanced  Course,  and  are 
called  imaginary  numbers.  For  us  any  quadratic  equation  which  gives 
rise  to  such  a  solution  is  to  be  interpreted  as  stating  some  impossible 
condition. 

EXERCISES 

1.  7-3^  =  5  x2.  4.   12 -51  a  =  36  +  6^. 

2.  51^-33  =  3^.  5.   5aj  +  a?  +  8  =  0. 

3.  Ux  +  8-x*=52-3x2.  6.   5x*-31x=-§. 


QUADRATIC  EQUATIONS  245 

7.  x2-8  +  3a;=-15a;.  10.    37  -  4  x2-  12x  =  79  -5a* 

8.  llar9-49a:  +  57  =  0.  11.    10  a? +  41  +  7  a:  =  44. 

9.  3a^  +  18-16x=5.  12.   45  +  3^-85-2^  =  0. 

MISCELLANEOUS  QUADRATICS 

Solve  as  many  as  possible  of  the  following  equations  by  fac- 
toring. When  this  is  not  convenient,  use  the  formula  of  §  179, 
or  complete  the  square  independently  in  each  case.  Show 
which  equations  state  impossible  conditions.  Approximate  all 
square  roots  to  two  decimal  places. 

1.  x2  + 11  x  =  210.  21.  2a:2  +  3a;-3  =  12a;+2. 

2.  5^-3^  =  4.  22.  3^-7^  =  10. 

3.  7x  +  3  x2  -18  =  0.  23.  17a  +  31  +  2ar°  =  0. 

4.  2=  5  a; +  7  a*  24.  18  -  41  a  =  3  +  a* 

5.  6z-lla:2  =  -7.  25.  10 x  +  25  =  5 - 2 x -x2. 

6.  -51+42a;-3ar!  =  0.  26.  3a-59  +  ar>  =  0. 

7.  3ar!  +  3a:  =  2x-  +  4.  27.  5ar+7z-6  =  0. 

8.  13-8a  +  3ar  =  0.  28.  x>+A2=lx. 

9.  2^ +  11  a=32  a; -a?- 27.    29.  8  a; -5  a?2  =  2. 

10.  176  +  3a;-a;2  =  2x.  30.  5^  +  3^-22  =  0. 

11.  x2  +  6x-54  =  0.  31.  50  +20 ^  +  ^  =  5 x. 

12.  5  x2  +  9  x  +  12  =  4  x2  +  a;.     32.  x2  +  x  +  4  =  0. 

13.  2a*-4x-25  =  0.  33.  20  05+2  3?  + 42=33  a>  + a* 

14.  7.^  +  11  z  =  6.  34.  17x-3x-2=-6. 

15.  2z2-ll£  +  5  =  0.  35.  8a?  +  5a?2=-2. 

16.  2ar!-lla;  =  6.  36.  10  + 15 x  +  a;2  =  26 x. 

17.  25  x  -  95  =  x2.  37.  3  x2  -  2x  -7  =  0. 

18.  11  x2 - 42z  =  2.  38.  5x2-9aj-18  =  0. 

19.  a;2- 8  a;- 4  =  a?-22.  39.  7  a; -7  a;2 +  24  =  0. 

20.  8a;2  +  5a;  =  -8.  40.  31 +2  a?  +  a;2  =  0. 

41.   7z2  +  7a:-5a;2  +  20  =  ar2-2a;  +  2. 


246  QUOTIENTS  AND   SQUARE  BOOTS 

181.  The  solution  of  two  equations  in  two  variables,  one  of 
which  is  linear  and  the  other  quadratic,  can  be  reduced  to  the 
solution  of  a  quadratic  equation  in  one  variable.     (See  §  149.) 

Example.    Solve        f      x  +  y  =  3,  (1) 

(3x2-y2  =  U.  (2) 

From  (1),  y  =  3-x.  (3) 

Substituting  in  (2)  and  reducing, 

2x2+6x-23  =  0.  (4) 

Substituting  in  formula  §  179,  *  =  -  6  ±  V36  -  4 -2(- 23)^  _ 

4 
Hence  xl  =  2.21  and  x2  =  —  5.21. 

Substituting  these  values  of  x  in  (1)  we  have  as  the  approximate 
roots, 


=  2.21 1 
=  0.79  J 


yi  =  0.79  J  amS2  =  8.21      | 

yx  and  y2  are  here  used  to  designate  the  values  of  y  which  correspond 
to  xx  and  x2  respectively. 

In  this  manner  solve  the  following  equations  simultaneously, 
finding  in  each  case  two  pairs  of  roots.  In  the  case  of  roots 
which  are  neither  integers  nor  exact  fractions,  find  the  approxi- 
mate results  to  two  places  of  decimals. 

f*-y=l.  K_2/!  =  3 

'    lar,  +  w8  =  13. 


#2  =  13.  6-      9      4 

[x  —  y  =  4. 
x  +  y=9.  J 

^  +  2/2  =  41.  [x-y  =  l. 

3-   W  =  42.  l»     16 

.     j3x-y  =  5.  o    \2x+y  =  5. 

fa:  +  4y  =  26.  f*-y»a 

'  laj,-«'  =  ll.  '  1^-3^  =  13. 


!: 


QUADRATIC  EQUATIONS  247 

10.    J*-3^1-  16.    J*  +  2/=9- 

f3a-4</  =  l.  ly-2x  =  5. 

*   1  ic2  —  y2  =  24.  '   l?/2-3a;y  =  16. 

fz  +  2/  =  4.  f22/-3a,-  =  0. 

'  l2^-3^4-2/2  =  8.  '  l?/2  +  a2  =  52. 

13    J*-*-*-  19    \y-2x  =  5. 

\±xt  +  2xy-y2  =  19.  '   larJ  +  y2  =  40. 

f5*  +  y  =  12.  2Q    f*-4y  =  12. 

12^-3^  +  ^  =  0.  I3a2  +  2a:y-6y  =  44. 

|x-22/  =  3.  Jy-3a  =  f. 

'   l2y2-a2  =  4.  "    12^-3^  =  4. 


PROBLEMS 

In  each  problem  find  the  two  roots  of  the  quadratic  equation 
and  determine  whether  both  are  applicable  to  the  problem : 

1.  The  area  of  a  window  is  2016  square  inches  and  the  length 
of  the  frame  is  180  inches.    Find  the  dimensions  of  the  window. 

2.  The  area  of  a  rectangular  city  block,  including  the  side- 
walk, is  19,200  square  yards.  The  length  of  the  sidewalk  when 
measured  on  the  side  next  the  street  is  560  yards.  Find  the 
dimensions  of  the  block. 

3.  A  farmer  starts  to  plow  around  a  rectangular  field  which 
contains  48  acres.  The  length  of  the  first  furrow  is  376  rods. 
Find  the  dimensions  of  the  field. 

4.  A  rectangular  blackboard  contains  38  square  feet  and 
the  length  of  the  molding  is  27  feet.  Find  the  dimensions  of 
the  board. 

5.  A  park  is  120  rods  long  and  80  rods  wide.  It  is  decided 
to  double  the  area  of  the  park,  still  keeping  it  rectangular,  by 
adding  strips  of  equal  width  to  one  end  and  one  side.  Find 
the  width  of  the  strips. 


248  QUOTIENTS  AND   SQUARE  ROOTS 

6.  A  fancy  quilt  is  72  inches  long  and  56  inches  wide. 
It  is  decided  to  increase  its  area  10  square  feet  by  adding  a 
border.     Find  the  width  of  the  border. 

7.  A  city  block  is  400  by  480  feet  when  measured  to  the  outer 
edge  of  the  sidewalk.  At  4  cents  per  square  foot  it  costs  $  416.64 
to  lay  a  sidewalk  around  the  block.    Find  the  width  of  the  walk. 

8.  A  farmer  starts  cutting  grain  around  a  field  120  rods 
long  and  70  rods  wide.  How  wide  a  strip  must  he  cut  to 
make  12  acres  ? 

9.  The  sides  of  a  right  triangle  are  6  and  8  inches  respec- 
tively. How  much  must  be  added  to  each  side  so  as  to  increase 
the  hypotenuse  10  inches  ? 

10.  A  rectangular  lot  is  16  by  12  rods.  How  wide  a  strip 
must  be  added  to  one  end  and  one  side  to  obtain  a  rectangular 
lot  whose  diagonal  is  1  rod  greater  ? 

11.  A  picture  is  15  inches  by  20  inches.  How  wide  a  frame 
must  be  added  to  increase  the  diagonal  3  inches  ? 

12.  An  athletic  field  is  800  feet  long  and  600  feet  wide. 
The  field  is  to  be  extended  by  the  same  amount  in  length  and 
width  so  that  the  longest  possible  straight  course  (the  diagonal) 
shall  be  increased  by  100  feet.  What  will  be  the  dimensions 
of  the  new  field  ? 

13.  A  starts  north  from  a  certain  place  going  4  miles  per 
hour  and  B  starts  east  from  the  same  place  at  the  same  time 
going  3  miles  per  hour.  In  how  many  hours  will  they  be 
16  miles  apart, the  earth's  surface  being  considered  as  a  plane? 

Let  t  equal  the  required  number  of  hours. 
Then        (4  02  +  (3/)2  =  162  =  256. 
16  P  +  9  fi  =  256, 
25  fi  =  256. 
5t=  ±16. 
t  =  ±  3£. 

The  solution  t  =  —  3£  is  not  applicable  to  this 
problem. 


QUADRATIC  EQUATIONS 


249 


14.  In  the  preceding  problem  if  A  goes  5  miles  per  hour  and 
B  4  miles  per  hour,  in  how  many  hours  will  they  be  24  miles 
apart  ? 

15.  A  rectangle  is  12  inches  wide  and  16  inches  long.  How 
much  must  be  added  to  the  length  to  increase  the  diagonal 
4  inches  ? 


Let  x  =  number  of  inches  to  be  added  to  the  length.  The  diagonal 
of  the  original  rectangle  is  Vl22  +  162  =  20.  Hence  the  diagonal  of 
the  required  rectangle  is  24. 

Then  122  +  (16  +  x)2  =  242, 

or  x2  +  32  x  -  176  =  0. 

Solving,  x,  =  -  16  +  12  V3  =  4.78, 

and  x2=  -  16-12  V3  =  -36.78. 

The  negative  solution  obtained  here  may  be  taken  to  mean  that  if 
the  rectangle  is  extended  in  the  opposite  direction  from  the  fixed 
corner,  we  shall  get  a  rectangle  which  has  the  required  diagonal.  See 
the  figure. 

16.  How  much  must  the  width  of  the  rectangle  in  problem 
15  be  extended  so  as  to  increase  the  diagonal  by  4  ? 

17.  A  trunk  30  inches  long  is  just  large  enough  to  permit  an 
umbrella  36  inches  long  to  lie  diagonally  on  the  bottom.  How 
much  must  the  length  of  the  trunk  be  increased  if  it  is  to 
accommodate  a  gun  4  inches  longer  than  the  umbrella? 

18.  A  rectangle  is  21  inches  long  and  20  inches  wide.  The 
length  of  the  rectangle  is  decreased  twice  as  much  as  the 
width,  thereby  decreasing  the  length  of  the  diagonal  4  inches. 
Find  the  dimensions  of  the  new  rectangle. 


250  QUOTIENTS  AND  SQUARE  ROOTS 

19.  In  a  rectangular  table  cover  24  by  30  inches  there  are  two 
strips  of  drawn  work  of  equal  width  running  at  right  angles 
through  the  center  of  the  piece.  What  is  the  width  of  these 
strips  if  the  drawn  work  covers  one-tenth  of  the  whole  piece  ? 

20.  A  certain  university  campus  is  100  rods  long  and  80 
rods  wide.  There  are  two  driveways  running  through  the 
center  of  the  campus  at  right  angles  to  each  other  and  parallel 
to  the  sides.  What  is  the  width  of  these  driveways  if  their 
combined  area  is  356  square  rods  ? 

21.  A  farm  is  320  rods  long  and  280  rods  wide.  There  is  a 
road  2  rods  wide  running  around  the  boundary  of  the  farm  and 
lying  entirely  within  it.  There  is  also  a  road  2  rods  wide 
running  across  the  farm  parallel  to  the  ends.  What  is  the  area 
of  the  farm  exclusive  of  the  roads  ? 

22.  A  rectangular  park  is  480  rods  long  and  360  rods  wide. 
A  walk  is  laid  out  completely  around  the  park  and  a  drive 
through  the  length  of  the  park  parallel  to  the  sides.  What  is 
the  width  of  the  walk  if  the  drive  is  three  times  as  wide  as 
the  walk  and  the  combined  area  of  the  walk  and  the  drive  is 
3110  square  rods  ? 

23.  The  sum  of  the  sides  of  a  right  triangle  is  18  and  the 
length  of  the  hypotenuse  is  16.     Find  the  length  of  each  side. 

24.  The  length  of  a  fence  around  a  rectangular  athletic  field 
is  1400  feet,  and  the  longest  straight  track  possible  on  the  field 
is  500  feet.     Find  the  dimensions  of  the  field. 

Using  100  feet  for  the  unit  of  measure  the  equations  are 

ix  +  y  =  7, 
\x*  +  y*  =  25. 

25.  The  difference  between  the  sides  of  a  right  triangle  is  8 
and  the  hypotenuse  is  42.     Find  the  lengths  of  the  sides. 

26.  A.  room  is  5  feet  longer  than  it  is  wide  and  the  distance 
between  two  opposite  corners  is  25  feet.  Find  the  length  and 
width  of  the  room. 


QUADRATIC  EQUATIONS  251 

27.  One  side  of  a  right  triangle  is  8  feet,  and  the  hypotenuse 
is  2  feet  more  than  twice  the  other  side.  Find  the  length  of  its 
hypotenuse  and  of  the  remaining  side. 

28.  A  vacant  corner  lot  has  a  50-foot  frontage  on  one  street. 
What  is  the  frontage  on  the  other  street  if  the  distance  between 
opposite  corners  along  the  diagonal  is  110  feet  less  than  twice 
this  frontage. 

In  an  old  Chinese  arithmetic  said  to  have  been  written  about  2600 
B.C.,  we  find  the  following  two  problems  in  each  of  which  the  square 
of  the  unknown  occurs  but  cancels. 

29.  In  the  middle  of  a  square  pond  whose  sides  are  10  feet 
there  grows  a  reed  which  reaches  1  foot  above  the  water. 
When  the  reed  is  bent  over  to  the  side  of  the  pond,  it  just 
reaches  the  top  of  the  water.     How  deep  is  the  water  ? 

30.  A  bamboo  reed  10  feet  high  is  broken  off  so  that  the  top 
reaches  the  ground  just  three  feet  from  its  base.  How  far 
from  the  ground  is  the  reed  broken  off  ? 

31.  The  sum  of  the  squares  of  two  consecutive  integers  is 
13,945.     Find  the  numbers. 

32.  The  product  of  two  consecutive  integers  is  4422.  Find 
the  numbers. 

33.  A  square  piece  of  tin  is  made  into  an  open  box,  contain- 
ing 864  cubic  inches,  by  cutting  out  a  6-inch  square  from  each 
corner  of  the  tin  and  then  turning  up  the  sides.  Find  the 
dimensions  of  the  original  piece  of  tin. 

34.  A  rectangular  piece  of  tin  is  8  inches  longer  than  it  is 
wide.  By  cutting  out  a  7-inch  square  from  each  corner  and 
turning  up  the  sides,  an  open  box  containing  1260  cubic  inches 
is  formed.     Find  the  dimensions  of  the  original  piece  of  tin. 

35.  By  cutting  out  a  square  8  inches  on  a  side  from  each 
corner  of  a  sheet  of  metal  and  turning  up  the  sides,  we  obtain 
an  open  box  such  that  the  area  of  the  sides  and  ends  is  4  times 
the  area  of  the  bottom.  Find  the  dimensions  of  the  original 
sheet  if  it  is  twice  as  long  as  it  is  wide. 


252 


QUOTIENTS  AND   SQUARE  BOOTS 


36.  An  open  box  whose  bottom  is  a  square  has  a  lateral  area 
which  is  400  square  inches  more  than  the  area  of  the  bottom. 
Find  the  other  dimensions  of  the  box  if  it  is  10  inches  high. 
(By  lateral  area  is  meant  the  sum  of  the  areas  of  the  four 
sides.) 

37.  A  box  whose  bottom  is  4  times  as  long  as  it  is  wide 
has  a  lateral  area  600  square  inches  less  than  4  times  the  area 
of  the  bottom.  Find  the  dimensions  of  the  bottom  if  the  box 
is  6  inches  high. 

38.  A  train  approaching  Chicago  from  the  south  at  the 
rate  of  50  miles  per  hour  is  75  miles  away  when  a  train 
starts  west  from  Chicago  at  the  rate  of  25  miles  per  hour. 
How  long  after  the  second  train  starts  will  they  be  50  miles 
apart  measured  diagonally  across  the  country  ? 

If  t  is  the  number  of  hours  required,  then  (75  -  50 1)2  +  (25  <)2=  (50)2. 
This  may  be  written  :  (25)2  •  (3  -2  02+  (25)2  •  P  =  22  -  (25)2. 

Hence  dividing  both  members  by  (25) 2,  we  have  (3  —  2  t)2  +  t2  =  4. 

39.  An  automobile  run- 
ning northward  at  the  rate 
of  15  miles  per  hour  is  20 
miles  south  of  the  inter- 
section with  an  east  and 
west  road.  At  the  same 
time  another  automobile 
running  westward  on  the 
cross-road  at  the  rate  of  20 
miles  per  hour  is  15  miles 
east  of  the  crossing.  How 
far  apart  (diagonally)  will 
they  be  15  minutes  later  ? 
One  hour  later  ? 

40.  Under  the  condi- 
tions of  problem  89  how 
long  after  the  time  speci- 


4/       C 

/                   CO 

/       <± 

/       15-3  n 

15 

s  J 

15-r2  / 

QUADRATIC  EQUATIONS  253 

fied  will  the  automobiles  be  10  miles  apart?     Is  there  more 
than  one  such  position? 

41.  What  are  the  rates  of  motion  of  the  automobiles  in 
problem  39  if  one  hour  later  they  are  5  miles  apart  and  3 
hours  later  they  are  35  miles  apart?     (See  the  figure.) 

If  the  rates  of  the  automobiles  are  rx  and  r2,  then  after  1  hour  we 

have  (20-r1)2  +  (15-r2)2  =  52,  (1) 

and  after  3  hours  we  have    (20  -  3  r,) 2  +  (15  -  3  r2)2  =  352.  (2) 

Simplifying  (1)  and  (2),  r*  +  r22  -  40  rx  -  30  r2  +  600  =  0,  (3) 

and  9  rt2  +  9  r22  -  120  rx  -  90  r2  -600  =  0.  (4) 

Multiplying  (3)  by  9,  subtracting  from  (4),  and  simplifying, 

4  rt  +  3  r2  =  100.  (5) 

The  solution  of  this  problem  may  now  be  completed  by  solving  (5) 
and  (1)  simultaneously. 

Note.  In  equation  (2),  20  —  3  r±  and  15  —  3  r2  are  both  negative 
numbers.  This  means  that  in  this  problem  the  distances  north  and 
west  of  the  crossing  are  negative,  while  those  south  and  east  are 
positive. 

REVIEW  QUESTIONS 

1.  Define  exponent.     Explain  the   difference  between  an 
exponent  and  a  coefficient. 

2.  Explain  why  and  under  what  circumstances  exponents 
are  added  in  multiplication. 

Show  that  (a2)3  =  a8,  (a3)4  =  a12. 

3.  Explain  the  method  of  multiplying  two  monomials. 
How  is  Principle  III  involved  in  Principle  XV  ? 

4.  What  is  meant  by  factoring  ?     Is  the  following  expres- 
sion factored  ?    x  (a  +  b)  +  y  (a  +  b).     Why  ? 

5.  What  are  the  characteristics  of  a  trinomial  square  ? 
Are   the  following  trinomials  squares  ?     If  not,  change  one 

term  in  each  so  as  to  make  it  a  square,     x2  +  xy  +  y2 ;  x*  +  x*y2 
+  y*.  a2-2ab-b2;  4a2  +  4a6  +  462. 


254  QUOTIENTS  AND  SQUARE  ROOTS 

6.  What  are  the  factors  of  the  difference  of  two  squares  ? 
Factor  x6  —  ?/6  as  the  difference  of  two  squares. 

7.  What  are  factors  of  the  difference  of  two  cubes  ? 
Factor  x6  —  y6  as  the  difference  of  two  cubes. 

8.  What  are  the  factors  of  the  sum  of  two  cubes  ? 
Factor  x6  +  y6  as  the  sum  of  two  cubes. 

9.  Explain  how  to   factor  a  trinomial  by  inspecting  the 
end  products  and  cross-products  of  two  binomials. 

By  this  method  factor, 

3^-7  a-10;  a2-9a:  +  18;  3^  +  5^-12. 

10.  By  means  of  the  following  examples  explain  the  pro- 
cess of  factoring  by  grouping. 

x2  +  ax  +  bx  +  ab ;  x3  —  x  —  3  x2  +  3. 

11.  How  may  a  quadratic  equation  be  solved  by  factoring  ? 
Why  is  it  essential  that  one  member  of  the  equation  be  zero 

when  the  other  member  is  factored  ? 

12.  Explain  the  method  of  solving  the  quadratic  equation 
by  completing  the  square. 

13.  How  many  roots  has  a  quadratic  equation  ?  When  does 
the  solution  of  a  quadratic  equation  indicate  that  the  condi- 
tions of  a  problem  are  impossible  ? 

14.  Explain  why  and  under  what  circumstances  exponents 
are  subtracted  in  division. 

15.  Explain  the  method  of  finding  the  quotient  of  two  mo- 
nomials.    Show  how  Principle  V  is  involved  in  Principle  XVII. 

16.  State  Principle  XVIII.  Given  V7  =  2.645,  find  V28  by 
means  of  this  principle. 

17.  Explain  how  a  number  in  Arabic  figures  is  divided  into 
groups  for  the  purpose  of  finding  its  square  root. 

18.  Show  how  the  value  of  the  following  may  be  approxi- 
mated by  finding  only  one  square  root. 

5  V20  +  2  V45  -  3  V80  +  2  VJ. 


CHAPTER   VIII 
LITERAL   FRACTIONS 

COMMON  FACTORS 

182.    If  a  number  is  a  factor  of  each  of  two  or  more  numbers, 
it  is  said  to  be  a  common  factor  of  these  numbers. 

Thus,  8  is  a  common  factor  of  16  and  48,  and  12  is  a  common 
factor  of  12,  36,  and  48. 

If  each  of  a  given  set  of  numbers  is  factored  into  prime  factors, 
any  common  factor  which  they  may  have  is  at  once  apparent. 

Illustrative  Example.     Find  the  common  factors  of  72,  96, 
120,  and  288. 

Factoring,  we  have 


72  =  2 

.  o 

.  2 

3 

•  3  =  2s  •  32. 

96  =  2 

■  2. 

,  2 . 

■  2 

•  2  •  3  =  26  .  3. 

120  =  2 

•  2 

•  2 

•  2 

•  3  .  5  =  24  •  3 

•5. 

288  =  2 

■  2 

•  2 

•  2 

•  2  •  3  •  3  =  26 

8*. 

The  common  prime  factors  are  2,  2,  2,  and  3.  The  other 
common  factors  are  the  various  combinations  of  these,  namely, 
2.2  =  4,2.2.2  =  8,2.3  =  6,4.3  =  12,  and  8-3  =  24.  24 
is,  therefore,  the  greatest  common  factor  of  these  numbers. 

To  find  common  factors  of  literal  expressions  we  proceed 
in  the  same  manner. 

Illustrative  Example.  Find  the  common  factors  of  x  +  y, 
x2  —  y2,  and  x2  +  2  xy  +  y2. 

Factoring,  we  have  x  +  y  =  x  +  y. 

x2  -y2  =  (x  +  y)(x  -  y). 
x2  +  2  xy  +  y2  =  (x  +  y)  {x  +  y) . 

Hence,  x  -f  y  is  the  only  common  factor  of  these  expressions. 

255 


256  LITERAL  FRACTIONS 

Illustrative  Example.     Find  the  common  factors  of 
10  x2  +  20  xy  + 10  y2,  5(x  +  y)(x>  -  y2),  and  15{x  +  y)(x*  +  f). 

Factoring,  10  x2  +  20  xy  +  10#2  =  2  •  5(x  +y  )  (x  +  y~). 

5(x  +  #)(x2  -  #2)  =  5(x  +  y)(*  +  #)(x  -  y). 
15(x  +  ^)(x8  +  ys)  =  3  •  5  (x  +  y)(x  +  ?/)(x2  -  xy  +  #2). 

The  common  prime  factors  are  5,  x  +  y,  and  a;  +  ?/.  The 
other  factors,  obtained  by  combining  these,  are  5(cc  +  y),  (x  +  y)2, 
and  5(x  +  y)2.  The  last  factor,  5(sc2  +  2a$  +  2/2),  is  called  the 
highest  common  factor. 

The  name  highest  instead  of  greatest  is  used  in  algebra  referring 
to  the  degree  of  the  factor.  Thus  x2  is  of  higher  degree  than  x,  al- 
though if  x  =  \,  x2  is  not  greater  than  x. 

183.  Definition.  In  general  that  common  factor  which  con- 
tains the  greatest  number  of  prime  factors  is  the  highest  common 
factor.     This  is  usually  abbreviated  to  H.  C.  F.     (See  §  130.) 

EXERCISES 

Find  the  H.  C.  F.  of  the  following  sets  of  expressions : 

1.  x-y,  x?-y2,  x2  —  2xy  +  tf. 

2.  x2  +  2x  +  l,3x  +  6x?  +  3xi. 

3.  x2  +  4  x  +  4,  x2  -  6  x  — 16. 

4.  x2-8x  +  16,x2  +  10x-56. 

5.  as-b3,  a2-2ab  +  b2. 

6.  xP+y3,  x2  —  y2,  x2  +  2xy-\-y2. 

7.  x2  —  7  x  + 12,  ax  -  3  a  —  bx  +  3  b. 

8.  a2-13a  +  42,  a3-216,  a2-a-30. 

9.  27  +  f,y2  +  9y  +  18,  f-9. 

10.   b2  +  7b-S0,  62+H6-42,  62-&-6. 


COMMON  MULTIPLES  257 

11.  a3  +  2a2  +  a,  a2  +  a,  a3 +  5  a2 +  4  a. 

12.  Xs  +  y3,  Xs  +  afy  +  #?/2  +  S/3. 

13.  a4  +  3  ar5  +  2  a2,  &  +  x2,  x*  +  7  a?+  6  x2. 

14.  ar2  — 11  x  -f  30,  xz  -  5  2  +  x2  —  5  x. 

15.  ra3  —  w3,  2  xfm2  4-  2  tfmn  +  2  a^w2. 
.    16.  x>-l,a?-  1,3^-13 x  +  12. 

17.  1  -  64  a,-3,  1  - 16  ar2,  5  -  2  2  -  20  x  +  8  a*. 

18.  1  + 125  a3,  1  +  10  a  +  25  a2,  1-25  a2. 

19.  ac  —  ax +  3  be  —  3bx,  a3+27&3. 

20.  5  c  —  2,  5  ac  +  20  c  —  2  a  -  8. 

21.  4x*  -x2,2xi  +  xi-x2,2x4-3xi  +  x2. 

22.  3a3-3a,  3a3-6  a2+-3a,  6a3  +  12a2-15a. 

23.  6  a;  — 10  xy  +  4  xy2,  18  a;  —  8  xy2,  54  a;  — 16  xy3. 

24.  3x5  +  9x4-3x3,5x2y2-{-  15xy2-5y%  7 ax2  +  21ax-7a. 

25.  18  3^-57  3^  + 30  a:,  9  a3  - 15  a:2  +  6  a;,  18  ar»  -  39  x2  + 18  a;. 

COMMON   MULTIPLES 
184.    A  number  is  said  to  be  a  multiple  of  any  of  its  factors. 
In  particular  any  number  is  a  multiple  of  itself  and  of  one. 

Thus,  18  is  a  multiple  of  1,  2,  3,  6,  9,  and  18,  but  not  of  12.  3  a2x2 
is  a  multiple  of  3,  3  x,  3  x2,  etc. 

Since  a  multiple  of  a  number  is  divisible  by  that  number,  it 
must  contain  as  a  factor  every  factor  of  that  number. 

E.g.  108  is  a  multiple  of  54  and  contains  as  factors  all  the  factors 
of  54,  namely  3,  3,  3,  and  2,  and  also  2,  6,  9,  18,  and  54. 

Definition.  A  number  is  a  common  multiple  of  two  or  more 
numbers  if  it  is  a  multiple  of  each  of  the  numbers. 

Thus,  18  is  a  common  multiple  of  6,  9  and  18.  Evidently  3  •  18, 
4  •  18,  5  •  18,  etc.  are  also  common  multiples  of  6,  9  and  18.  Of  all 
these  common  multiples  18  is  the  smallest  and  is  called  the  least  or 
lowest  common  multiple. 


258  LITERAL  FRACTIONS 

185.  The  process  of  finding  the  least  common  multiple  of  a 
set  of  numbers  in  Arabic  figures  is  shown  as  follows : 

Illustrative  Example.  Find  the  least  common  multiple  of 
16,  24,  and  98. 

Finding  the  prime  factors  of  each  number, 
16  =  2  •  2  •  2  •  2  =  2* 
24  =  2- 2. 2- 3  =  28- 3 
98  =  2  •  7  •  7      =  2  •  72 

Any  multiple  of  16  contains  all  its  prime  factors,  namely, 
2,  2,  2,  and  2.  Any  common  multiple  of  16  and  24  contains  in 
addition  to  the  prime  factors  of  16  any  prime  factors  of  24  not 
in  16,  namely  3.  Any  common  multiple  of  16, 24,  and  98  contains 
in  addition  to  2, 2, 2, 2  and  3  those  prime  factors  of  98  not  in  16 
or  24,  namely  7  and  7.  Hence  2- 2- 2- 2- 3-7-7  =  1176  is 
a  common  multiple  of  16,  24,  and  98.  Moreover,  it  is  the  least 
common  multiple  because  no  unnecessary  factor  has  been  included. 

We  proceed  with  a  set  of  literal  expressions  in  the  same 
manner  as  above. 

Illustrative  Example.    Find  the  L.  C.  M.  of 

ar2  —  2Z2;  x2  +  2 xy  +  y2 ;  x?-2xy  +  y\ 

Factoring,  x2  —  y*  =  (x  —  y)  (x  +  y).  (1) 

x*  +  2xy  +  y*=  (x  +  y)(x  +  y).  (2) 

x*-2xy  +  y*  =  (x-y)(x-y).  (3) 

In  order  that  an  expression  may  be  a  multiple  of  (1)  it  must  con- 
tain the  factors  x  —  y  and  x  +  y.  To  be  a  common  multiple  of  (1) 
and  (2)  it  must  contain  a  second  factor  x  +  y,  giving  (x  —  y),  (x+y), 
(x  +  y).  To  be  a  common  multiple  of  (1),  (2)  and  (3)  it  must 
contain  a  second  factor  x  —  y,  giving  (x  —  y),  (x  +  y)  ,  (x  +  y),  (x  —  y). 
The  product  thus  found  contains  the  fewest  prime  factors  possible 
in  order  to  be  a  common  multiple  of  (1),  (2),  and  (3).  Hence 
(x  —  y)(x  +  y)  (x  +  y)(x  —  y)  =  (x  —  y)2  (x  +  y)'2  is  called  the  lowest 
common  multiple  of  (1),  (2),  and  (3),  since  it  is  the  common 
multiple  of  lowest  degree. 


COMMON  MULTIPLES  259 

In  general,  the  process  may  be  described  as  follows :  to 
obtain  the  lowest  common  multiple  of  a  set  of  expressions, 
factor  each  expression  into  prime  factors;  use  all  factors  of 
the  first  expression  together  with  those  factors  of  the  second 
which  are  not  in  the  first,  those  of  the  third  which  are  not  in 
the  first  and  second,  etc.  It  is  evident  that  in  this  manner  we 
obtain  a  product  which  is  a  common  multiple  of  the  given 
expressions,  but  such  that  if  any  one  of  these  factors  is  omitted, 
it  will  cease  to  be  a  multiple  of  some  one  of  the  expressions ; 
that  is,  it  will  no  longer  be  a  common  multiple  of  them  all. 

Thus,  if  in  the  example  above  either  of  the  factors  x  —  y  is  omitted, 
the  product  will  no  longer  be  a  multiple  of  x2  —  2  xy  +  y2. 

186.   Definition.     We  now  define  the  lowest  common  multiple 

of  a  set  of  expressions  as  that  common  multiple  which  contains 
the  smallest  number  of  prime  factors.  The  lowest  common 
multiple  is  usually  abbreviated  to  L.  C.  M. 

EXERCISES 

Find  the  L.  C.  M.  of  the  following  expressions  : 

1.  2-3.4;  3-7-8;  2s-  3  -4.       9.    25^-1,  125^-1. 

2.  5«Y,  lOafy,  25ofy.  10.   2^-7  a+6,  4^-11  a>+6. 

3.  2  ab,  6  a2,  4  b2c.  11.   x^  —  y^jX  —  y^  +  xy-^y2. 

4.  x2  —  y2,  x2  —  2xy  +  y2.  12.   x3  —  y3,  Xs  +  y3,  x2  —  y2. 

5.  x-y^  +  y^tf  —  y2.  13.   5 ^+7^-6,  x2— 15a;—  34. 

6.  4  —  x2,  2  —  x,  2  +  x.  14.   cc3  +  2/3,  x2  —  y2,  (x—  y)2. 

7.  a2+2a&+&2,a2-2ao  +  o2.  15.' 3a6c,a2-4ac+4c2,  a-2c. 

8.  x2+  3 x  +  2,  x2- 4,^-1.  16.   x2-l,  x  +  1,  a2  +  8a  +  7. 

17.  4  x3y  -  44 x2y  + 120 xy,  3  a?x2  -  22  a?x  +  35  a3. 

18.  x2  +  2xy+y2,  2aa^— 10ax  +  12a. 


260  LITERAL  FRACTIONS 

19.  3  6a^-216a;  +  36  6,  ar*- 5a; +  4. 

20.  5a262-5aV,62  +  26c  +  6  +  c  +  c2. 

21.  15  (fax2  + 16  <?ax  +  (?a,  2  cax2  + 10  cax  +  8  ca. 

22.  ac  —  4  a  +  be  —  4  6,  aa:  -f  ac  +  6a;  +  6c. 

23.  a^  +  5a;  +  4,  a;2  +  7a;-f  12,a;2  +  5a;  +  6. 

24.  x2  +  2x+l,x2-2x  +  l,x2-l. 

REDUCTION  OF  FRACTIONS  TO  LOWEST   TERMS 
187.   In  arithmetic  the  denominator  of  a  fraction  is  usually- 
regarded  as  indicating  the  number  of  equal  parts  into  which  a 
unit  is  divided,  while  the  numerator  designates  a  certain  num- 
ber of  these  parts. 

Thus,  |  means  2  of  the  3  equal  parts  of  a  unit. 

5 
However,  a  fraction  such  as  —  cannot  be  regarded  in  this 

5 
way,  since  a  unit  cannot  be  divided  into  3^  equal  parts.     — 

5 
indicates  that  5  is  to  be  divided  by  3£.     I.e.  —  =  5  -*- 3\> 

In  algebra  any  fraction  is  usually  regarded  as  an  indicated 
division  in  which  the  numerator  is  the  dividend  and  the  de- 
nominator is  the  divisor. 

Thus,  -  is  understood  to  mean  a  ■+■  6. 
6 

The  numerator  and  denominator  are  together  called  the  terms 
of  the  fraction. 

In  case  the  numerator  and  denominator  have  common  factors, 
these  may  be  removed,  without  changing  the  value  of  the  frac- 
tion, by  means  of  Principle  XVII,  as  applied  in  §  157. 

Thus,  — - — '———  =  — ■ — — ,  where  the  common  factor  3  is  can- 
,         '  3  •  7  •  11        7  •  11  ' 
celed. 


REDUCTION   TO  LOWEST  TERMS  261 

Similarly  — - — ■ — -  =  - — -,  the  factors  23  3,  and  4  being  can- 

24  •  3  •  42       2  •  4 

celed;   and  <*"  7*  +  12>  =  (»-3)(«-*)  =  fez^. 
'  (^-Sa-t-e)       (a? -2)  (a? -3)      (a;-2) 

If  the  terms  of  a  fraction  have  no  common  factor,  the  frac- 
tion is  said  to  be  in  its  lowest  terms.  Evidently  the  process  of 
canceling  common  factors  in  the  numerator  and  denominator 
may  always  be  continued  until  the  fraction  is  reduced  to  its 
lowest  terms. 

EXERCISES 

Reduce  the  following  fractions  to  lowest  terms  : 

t     3 .  92 .  2G  aW  x*-f 

'  24.53-94'        '  a262'  '  2x2-3xy  +  y2' 

4  a  W  x*  +  2xy  +  y2  64  -ft3 

8aW*  "         tf-tf      '  '  16-86 +  &2' 

.    afyV  a    a:2  +  7a;-30  _  ar3  +  27z3 


a^V                    a?2  —  7  a;  + 12  xy  —  5  x+  3  yz  —  15  z 

10             1-216C3  27  •  3s .  54  -  26 .  34  •  57 

a;-4y-6ca;  +  24c2/'  '   24-32  •  55-25  •  32- 52' 

,j    14  6z  —  2  6a;  +  ax  —  7  az  5  x?y*  —  12a*Y  +  7  a3?/2 

ar!-4922  '           6  afy2  +  3  x?y2 

.„       3a2-29a  +  56  nQ    5c  +  106- 6c-262 

63 -9  a- 7m +  ma  8c3  +  64  63 

13  a(x  -  Vf  19    4  a;4 -28  3^  +  48  0^ 
(a?-tf)(x-y)  '     2x*-$a?  +  6x2  ' 

14  a;3 +27  20    9a264+ 18a263c+9a262c2 
4a2  +  24a;  +  36'  "              3a63-3a6c2 

15  a2-3a-36  +  a6  7 a?y2  - 133 xy  + 126 a; 

(a2-b2)(a-3)  '  15xy2-36xy +  21x  ' 


262  LITERAL  FRACTIONS 

22  20xi  +  20x2y  +  5xy2  (a?-l)(s-2)(a?-3)(s-4) 

.  Wtf-Wafy2  '   (a>-l)(a>-3)(a-3)(a;-4)' 

23  3a^-3a52c2  (s8-yg)(as»  +  2sy  +  y8) 
27  a362  +  27  a36c                        *    (a2  -  2  ^  +  ?/2)  (a?  +  y)  ' 

24  4a3-42a2  +  20a  (s»-l)(a3  +  l)(3s?  +  3) 

2a466-20a3&6  3(»4-l) 

28  (3  a*  ~  3  a2b2)  (a2  + 13  a  +  42) 

(a2  +  3a  +  2)(a2  +  5a-6)  ' 

29  23  •  43  •  54  (x2  -  b2)  (x2  + 19  x  +  90) 

'  22.4-53(«2  +  9a;-10)(ari  +  10a;  +  9)' 


REDUCTION  OP  FRACTIONS  TO  A  COMMON  DENOMINATOR 
188.  Since  we  have  just  seen  that  a  common  factor  may  be 
removed  from  the  numerator  and  denominator  of  a  fraction 
without  changing  the  value  of  the  fraction,  it  is  evident  that  a 
factor  may  be  introduced  into  both  terms  without  changing 
the  value  of  the  fraction. 

Thus  since  — ^—  is  reduced  to  -  by  removing  the  factor  3  from  both 
3  •  5  5 

4  3-4 

numerator  and  denominator,  so  -  is  changed  to by  introducing 

o  o  •  o 

the  factor  3   in  both  the  terms.     Likewise  any  other  factor  may  be 

introduced.      E.g.  *  =  ^  ,  ^  =  («-»)(«  +  »)  =  «^J*#    ' 
y    5      5-7     a  +  b      (a  +  b)(a  +  b)     (a  +  6)a 

In  this  manner  any  fraction  may  be  changed  into  an  equal 
fraction  whose  denominator  is  any  given  multiple  of  the  de- 
nominator of  the  given  fraction. 

Thus,  f  can  be  changed   into  a  fraction  whose  denominator  is 

q        o  _  -i  o 

72  (a  multiple  of  4)  by  multiplying  both  terms  by  18,  i.e.  '-.  —  j— ttt  • 


REDUCTION  TO  COMMON  DENOMINATOR  263 

and  can  be  changed  into  a  fraction  whose  denominator  is 

x-  y 

x2  —  y2  by  multiplying  both  of  its  terms  by  x  +  y,  i.e., 

Sa  -2b=(Sa-2b)(x  +  y) 
x  —  y  x2  —y2 

Any  two  or  more  fractions  may  therefore  be  changed  into 
respectively  equal  fractions  which  shall  have  a  common  de- 
nominator, namely,  a  common  multiple  of  the  denominators  of 
the  given  fractions. 

Illustrative  Example.     Keduce     ~    ,  .     „     ,    to  frac- 

as+1    x  —  1    x l  —  1 

tions  having  a  common  denominator. 

The  L.C.M.  of  the  denominators  is  (x  —  l)(x  +  1).  Multiply  the 
numerator  and  denominator  of  each  fraction  by  an  expression  which 
will  make  the  denominator  of  each  new  fraction  (x  —  l)(x  +  1). 

x-1  __  (x-  l)(x-  1)  _     x2  -  2  x  +  1 
x  +  1      (x  +  l)(x-l)      (z  +  l)(x-l)' 

X+1=(X+1)(X+1)  =  (X  +   1)2  . 

x-l      (x-l)(x  +  l)      (x  +  l)(x-l)  ' 

2x  +  3_         2x  +  3 
x2-l       (x  +  l)(x-l)" 

It  is  best  to  indicate  the  multiplication  in  the  common  denomina- 
tor, since  this  makes  it  more  easily  apparent  by  what  expression  the 
numerator  and  denominator  of  a  fraction  must  be  multiplied  in 
order  to  reduce  it  to  a  fraction  with  the  required  denominator. 

It  should  be  noticed  that  there  are  three  signs  in  connection 
with  a  fraction,  the  sign  of  the  fraction  itself,  the  sign  of  the 
numerator,  and  the  sign  of  the  denominator.  It  follows  from 
the  law  of  signs  in  division  (Principle  XII)  that  any  two  of 
these  signs  may  be  changed  simultaneously  without  changing 
the  value  of  the  fraction. 

-p  a  _  —  a —  a  _  _    a 

'9'  b~  -b~    '    b  -b 


Thus, 


264  LITERAL  FRACTIONS 

This  is  useful  in  cases  like  the  following: 

x  + 1       x  1 

Reduce  ,  — — -,  and to  fractions  having  a  com- 

1  —  x  x'-  —  1  x-\-l 

mon  denominator. 

x  +  1  _—x-l  _(x  +  l)(—x  —  l)  _—x*  —  2x  —  l 
1-x       x-1    '  '    (x  +  l)(x  —  l)   ''  x*—l 

X  X  -,         1  X  —  1  X—  1 

,  and 


tf-\      tf-l'  x  +  \      (aj  +  l)(a?-l)      x*-l 

EXERCISES 

Reduce  each  of  the  following  sets  of  fractions  to  equivalent 
fractions  having  a  common  denominator. 

1     ^+3  4  4         2         g  -  1      a?  +  l_ 

x—y'    x*  —  2xy  +  y'2  3  —  x'    x  +  l'    x  —  3 

.     a  — 1  a+1  c     a  +  &       a— 6  a 


a2-&2'    a2  +  2a6+62  '    b-a     (a  +  &)      a2-&2 

3-a  a;+4  .1  1 


^-9^+20'  7<e*-26a;--8         '    a2  — 3<e  — 4'    a;2  +  3a;  +  2' 
a+1  a— 1 


7. 


8. 


a2-2a&  +  &2'    a2  +  2a&+62 
1  1  1 

a3_&3>       6_a'       a2+a&  +  &2* 

a  & 


10. 


5a2-4a-12'    a2  +  4a-12'    a-2 
a;  — 2  a  +  2  cc-1 


^-Sa-G'    rf  +  lZx-lOtf    x*  +  VZx  +  18 


In  each  of  the  following  exercises  reduce  the  given  fractions 
to  a  common  denominator  and  check  by  substituting  a  con- 
venient number  for  each  letter,  taking  care  that  no  denomina- 
tor becomes  zero : 


REDUCTION   TO   COMMON  DENOMINATOR  265 

11  mr  d  16       *{T-t)        W(T-Q) 
m_l'    i  +  n'  '     w(Q-T)'   ~w(Q-t) 

12  Rr  1  17  P  d(V-w) 
(B  +  r)(m-l)'    B  +  r           '    W(p-2t+l)'       W-w 

CS         Br  +  Sr  +  SB    t.         V  V  1 


B  +  BS'  BSs         '       '     V-v'    V+v'    V'-v1 

.  .         a            b  ,  «       BA  1  1 

14.     ,     •  19. 


15. 


n—an—b  a  —  A'B-\-r'A  —  a 

W(T-Q)      V  9ft     B     r        1  Bx 

w((J  —  t)       "jo  x      y     x  —  y     x-\-y 


189.  Since  any  number  may  be  written  as  a  fraction  with 
the  denominator  1,  the  above  process  may  be  used  to  reduce 
an  integral  expression  to  the  form  of  a  fraction  having  any 
desired  denominator. 

Thus,     3=^5;      x-y=(x-Mxi-V,ete. 
5  x2  —  1 

It  is  sometimes  convenient  to  reduce  expressions,  some  of 
which  are  not  fractions,  to  the  form  of  fractions  having  a  com- 
mon denominator. 

Illustrative   Example.       Reduce   5  x,  ~    ',    —r^,   to 

ar  —  1         x  —  1 

fractions  having  a  common  denominator.     The  lowest  common 

denominator  is  ar2  —  1. 

„,.  5  x  (x2  —  1)     5  x8  —  5  x 

Thus,        5x=- — \ — ,      - — • — T~> 

x2  —  1  x2  —  1 

5x  —  1  _  5x  —  1 

x2  -  1  "  x2  - 1  ' 

2  x  —  y  _  (2  x  —  ?/)(x  +  1)  _2x2  +  2x  —  yx  —  y 
x-1    "   (x-l)(x+l)  ~  x2-l 


266  LITERAL  FRACTIONS 


EXERCISES 

Reduce  the  following  mixed  expressions  to  fractions  having 
a  common  denominator : 

-  5x— 3  oil       .     2     «+ 1 

1.  5,  a>-y,.  •        3.    1  +  a  +  a2,    — ^rr- 

ar  +  2  #?/  +  ?/J  a  —  1 

2.  3a~c,    !•=•,  2c  +  2>     4     *+«*+*•,    *±*. 

x—y        x+y  x  —  y 

5.    rf  —  xy  +  y2,   x*  —  y2, 


7.  3  a  —  2&  —  c, 

8.  x*-l,   a?-l, 

9.  x2  +  2z?/  +  2/2, 

10.    x  +  y,   x-y,    -      -*-,    : 

x?  +  y2     x-y 

In  each  of  the  following  reduce  the  expressions  to  the 
form  of  fractions  having  a  common  denominator  and  check 
the  results  by  substituting  convenient  numbers  for  the 
letters : 

„.      **    ,.  „.    SA.,B-r.  13.    r,   «±-(- 

a  — A  a  — A  2 

14.      TP(g-Q       sT}     8(t-q)m  u      H}    hd, 

w  2  Z> 


#  > 

x  +  y 

X 

y  . 

x—y 

x  +  y 

5 
a  —  &' 

2 
b-c 

x  +  1 
x-1 

1 

x  +  y' 

1 
1-x 

x-y 

x  +  1 

ADDITION  AND  SUBTRACTION  267 


ADDITION   AND   SUBTRACTION   OF   FRACTIONS 

190.  Fractions  having  a  common  denominator  may  be  added 
or  subtracted,  exactly  as  in  arithmetic,  by  adding  or  subtracting 
the  numerators  and  dividing  the  result  by  the  common 
denominator. 

For  we  have,  by  Principle  VI,  — - —  =  —  +  —  =  4  +  5.     Like- 

o     |     e         o         k 

wise  — ■ —  =  t  +  t  ,  the  division  being  indicated  in  this  case.     Hence, 
4         4     4  ° 

3      5     3  4-5 
reading  this  identity  from  right  to   left,   we  have    ■?  +  r  =  — -z — . 

ti       •  6  4     _6-4         2         T 

Likewise — = .    In  general, 

x  —  y      x  —  y     x  —  y     x  —  y 

a     b  _a  +  b        ,   a     b_a  —  b 
c     c~     c    '  c     c~     c 

If  fractions  which  are  to  be  added  or  subtracted  do  not  have 
a  common  denominator,  they  must  be  reduced  to  this  form 

Example.    Add      ^  and  a2  +  2ah  +  **. 

a  +  b  a2  -  2  ab  +  b2 


Reducing  the  fractions  to  the  common  denominator 

(a-b)(a  -  b)(a  +  V), 
we  have 

a-b  __  (a-b)(a-b)(a-b)  _  a8-  3a2&  +  3  ab2-  b* } 
a  +  b      (a  +  b)(a  -  6)(a  -b)      (a +  b) (a-b) (a-b) 

&nd  a2+2ab  +  b*  _  (a  +  b) (a2  +  2  ab  +  b2)  _  a8+  3  a2b  +  3  ab2  +  bs 
a2-2ab  +  b2      (a  +  b)(a  -  b)(a  -  b)  ~  (a+b)(a  -  b)(a- b)' 

Adding  the  numerators,  we  have  2  a8  +  6  ab2 ;  whence  the  sum  of 

the  fractions  is 

2  a8  +  6  ab2 

(a  +  b)(a-  b)(a-b)' 


268  LITERAL  FRACTIONS 

EXERCISES 

Perform  the  following  additions  and  subtractions  : 

1.   3  |  2  I    a  +  b  8       2x*  y  y 

4     7     3  a— b  x?  —  y2     x  —  y     x  +  y 

x-y        x-y  a  +  1  q-1 


(x  +  y)2     tf  —  y2  tf  +  a  +  l     a2  — a  +  1 

3     ^-9^  +  18    |      x  1Q     _&_,       &  b2 


or2 -13  a; +  36     4 -a  l_&'i  +  &2     1_&2 

4    2  I      CT      I      &  11    a^  +  1  ■  a  +  1  ,  3a+2 

3     a  +  6     a-&  *   x-2     x  +  2      x2  —  ! 


3  5  2  ..    »-l      aj+1  ,  ar>-5 


23-32     22-34     24-33  a  +  1  a?-l  a2-l 

6          a2  —  9  &2          a2  —  6  q&  y2  y  y 

a2+6&  +  9&2     a2-962'  '   y2-l  y  +  1  l-y' 

7.   ^±1  +  ^^-2.  14.   i-I L.  +  ^l. 

»— y     »  +  y  «     y  x—y  x+y 

15    I         ^      .     2  a2  —  a 
2     6_a"l"&2_a2  +  &  +  a- 


16    ^  +  4^ 


a^  +  y3       »  +  y     tf  —  xy  +  y 


W.    .-JL-.  +  :    *  * 


1  — ar*     1  — a;     1  +  aj  +  a^ 


18.  a~3 CT~1 


q2-3a  +  2     a2_5a  +  6     o2-4a  +  3 


ADDITION  AND   SUBTRACTION  269 


19-      .        *       ..+      2 


20. 


a^-5a;-14      tc-7      a,-2 -9  a; +  14 

a & 

ac  +  ad—bc  —  bd     a?  —  2  a&  +  &2 


,!.       »     +.,39  x.    3 


a-5      a2  +  3a-40     a  +  8 


22.  ; " 1 "  „-- ^,+   4 


2/2  +  82/+16     y(y  +  4)     y2(y+4) 


23. = —  + 


a*  +  4:X-60     v*-±x-12 


n.      a—c    ,         c  —  a  2 

24.    -= „  + 


25. 


a2  —  c2     a2  +  2  ac  +  c2     a  —  c 

_9 8 

!  +  7a;-18      ^  +  6^-16' 


26.        a  +  2       |         «~4  «  +  2 


a2-a-6     a2-7a  +  12     a2-2a-8 


27.  *^_       *       +       1 


^-11*5  +  30     ^-36      a^-25 

(z-l)(a  +  2)  +  (a;  +  2)(a-3)+(a;-l)(3-z)' 

29  1 1  ,  1 

(«-&)(&-c)      (&-a)(c-d)      (b-c)(c-d) 


30        4  a-1  a2-38a-3 

'a-3     a2  +  3a  +  9  a3 -27 


270  LITERAL  FRACTIONS 


MULTIPLICATION  AND   DIVISION  OF  FRACTIONS 

191.  The  product  of  a  fraction  and  an  integer.  In  arithmetic 
the  product  of  a  fraction  and  an  integer  is  obtained  by  multi- 
plying the  numerator  of  the  fraction  by  the  integer. 

Thus,  2  x  4  =  —  and  7  x  -  =  — • 
3  3  5       5 

Since  in  algebra  a  fraction  is  an  indicated  quotient,  and  since 
multiplying  the  dividend  multiplies  the  quotient,  it  follows  that 
in  algebra  also  the  product  of  a  fraction  and  an  integral  expres- 
sion is  obtained  by  multiplying  the  numerator  by  the  integral 
expression. 

That  is,  in  general,  a  •  -  = =  — 

c        c         c 

It  is  best  to  factor  completely  the  expressions  to  be  multi- 
plied and  keep  them  in  the  factored  form  until  all  possible 
cancellations  have  been  made. 


EXERCISES 

Find  the  following  indicated  products  and  reduce  the  frac- 
tions to  the  simplest  form. 

1.    (l-o)  x1  +  a  +  a\  6.  <V+9a;+18)x       X~ 


a  —  1  x2—2x  —  15 

2.  (a3-?/3)  X  <^ty.  7.  (1  _^)  x  -   1~X     • 

x-y  v  J      1+x  +  a? 

3.  (a»-2«a  +  a!)x^         8.  (27 a3 -  1)  x  — ~^~. r- 

x  —  a  9or+3a  +  l 

5-  Jfrax(o!-5a+4>- 10-  *-****t&    . 


MULTIPLICATION  AND  DIVISION  271 

192.  In  arithmetic  a  fraction  is  divided  by  an  integer  by 
multiplying  its  denominator  or  dividing  its  numerator  by  the 
integer. 

—    ***-&  ¥^fH- 

Since  multiplying  the  divisor  or  dividing  the  dividend  divides 
the  quotient,  it  follows  that  an  algebraic  fraction  is  divided  by 
an  integral  expression  by  dividing  the  numerator  or  multiply- 
ing the  denominator  by  the  integral  expression. 

That  is,  in  general,    -  ■+■  c  =  —  =  — j- — 
b  be         b 


EXERCISES 

Find  the  following  indicated  quotients  and  reduce  the  frac- 
tions to  their  lowest  terms : 

,      x3  —  w3      /  o  ,         .     2\  n     #24-4#  +  4      ,  j,      .. 

1.    ?--i-  (ar  +  xy  +  y2).  3.      „„ — — — 5-(ar  — 4). 

z  +  y       v  *     * }  x2-2x-\-l      v  ; 

2.  t±t+(x>-y*).  4.    ^^^(1  +  3^+9^. 

a  — ?/  1  —  9ar* 

2^-35  ^_^_4a._5)> 


a?  +  10  x  +  21 


6.  ^7  ^  X  +,3r9  -*-  (s2  -  x  -  156). 

or  —  8  a;  -f- 15 

7.  ^  +  6c-ad-M  c_d 

xy  —  4:X—3y  +  12      x  ' 


x2  +  ax  +  bx  +  a6 
a2 -fax  —  3#—  3  6 


(cc2  +  aa?  —  5  x  —  5  a). 


9.    : -s-  (3  r  —  xr  +  3  s  —  xs). 

mx  —  m—nx  +  n 

10-      ^~Q3;~99^(a:2+9a;+2)- 

ar  —  9  x  —  22 


272  LITERAL  FRACTIONS 


TO   MULTIPLY   A   FRACTION   BY   A   FRACTION 

193.  In  arithmetic,  to  multiply  a  number  by  the  quotient  of 
two  numbers  is  the  same  as  to  multiply  by  the  dividend  and 
then  divide  the  product  by  the  divisor. 

E.g.  10  •  —  =  10  •  6  =  60;  or,  10  •  —  =  (10  •  18)  +  3  =60. 
2     5_/2.*\  .  „_»«6     -     2-5 


Likewise,  |  •  -  =  U  -5J--7 


3  3-7 


Since  in  algebra  a  fraction  is  an  indicated  quotient,  to  mul- 
tiply a  number  by  an  algebraic  fraction  we  multiply  by  the 
numerator  and  divide  the  product  by  the  denominator. 

rru  fl    o     fa      \      .     ac  .    .     ac 

Thus-     r^-fr-v*    *"     w 

Hence,  the  product  of  two  algebraic  fractions  is  a  fraction 
whose  numerator  is  the  product  of  the  given  numerators  and 
whose  denominator  is  the  product  of  the  given  denominators. 

Illustrative  Example.  Multiply  ^^  by  £=§*±f 
and  reduce  the  resulting  fraction  to  its  lowest  terms. 

x2-  1  ^-3x+2^(g-l)(ic  +  l)(x-2)(x-  1) 

x2-7x  +  10      x2  +  2x+l      (x  -  2)(x  -  5)(x  +  l)(*  +  1) 

_(x-  l)(x  -  1)  _  x2  -  2  x  +  1 
(x-5)(x+l)      x2-4x-5* 

It  is  desirable  to  resolve  each  numerator  and  denominator 
into  prime  factors,  and  then  cancel  all  common  factors  before 
performing  any  multiplication. 


MULTIPLICATION  AND  DIVISION  273 


EXERCISES 

Find  the  following  indicated  products  and  reduce  each  frac- 
tion to  its  lowest  terms : 

1    3  arfy2     6  az  32 .  43     10-2 

2yz2      9x*'  52 .  24         34 

5o(q-6)       9(a  +  &)2  6    32 .  23     54.  72      6  •  32 

3  c  (a +  6)      15  (a2- 62)  '      52        34  •  23     54  •  73 

12<*b         35  (c2+c5  + ft2) .  7    fl2-^     2^  +  43;  +  2 
'   5(c3-63)  14c*&»         '     '   x2-!         3a?  +  6x 

4    y»  +  3y  +  2     y»-7y  +  12     g    a2-10q+16      a  +  3 
'  y«_5y  +  6      f  +  Sy  +  7       '     a2+6a  +  9        a2-4' 

9#   (x  +  yf-z2  x  a;  x  (a;  -  y)2  -  z\ 

x2  +  xy  —  xz      (x  —  zf—y*     xy  —  y2  —  yz 

10     a2  +  7a  +  12      3a3  +  27a2  +  42a 
a3  +  5a2  +  6a      6a2  +  66a  +  168  ' 

3<(a  -  5)(a  +  2)     22(a  -  3)(a  -  5) 
'  23(a-3)(a-2)  33(a-5)2 


12. 


3(g  +  4)2  Q-7)2 

4(«  +  4)(as-7)      3(a>  +  4)(a>-7)' 


13  a3+5a2-36a     (q-16)(a-3) 

a2-7a-144      a(a-4)(a  +  2)" 

14  3a(a-7)(a-5)      6(a-3)(a+10) 
7  6(a-3)(a-7)     a(a-5)(a-10) 

15  42(6-3)0-4)      6(6  +  7)0-5) 

3(6-4)(6+7)       14(6-3)(6-6) 

a(»-<*)        (6»-q»)»        (6  +  a)2 

6(6  + a)       b2  +  ba  +  a2     (6  -  a)2 


274  LITERAL  FRACTIONS 

17  a2-4g  +  3      a2- 9a +  20       a2 -la 
a2  -  5  a  +  4     a2  -  10  a  +  21  X  a2  -  5  a' 

18  3  a(a  -  7)(o  -  5)     6(a  -  3)(o  +  10) 

'  7  6(a-3)(a-7)     a(a-5)(a- 10)' 

19  a2- 12a +  35        a2 -8a +  15 

'  c(a2- 10a +  21)     a2  +  3a-70' 

20  c*-10c  +  21      c^-3c-88 

c2- 18  c +  77     c2-8c  +  15' 

ar^ar8  -  16)  y2  -  12  y  +  35     y«(g  +  5) 

'  y2(y2 -3 y- 28)      a?-9x+2Q      a?(x-4) 


22. 


3 a262(c2-14 c+33)       2(c2  +  c  -  56)  c2-2c-15 

4  a*  (c2  - 10  c  +  21)      a2(c2  - 16  c  +  55)      62(c2 + 12  c + 32) ' 


St2-2t-l     2t2  +  5t-S     4t2  +  10t  +  4, 
2t2  +  t-l      3t2  +  7t  +  2      4:t2-2t-2' 

m.     6^-7^  +  2  „6a?-5x-l      10a^  +  3a;-l 
'  10a^-7a;  +  l       6^+a-l        Ztf-kx-l 

452-175+4     1052-216  +  9     362-5  5  +  2 
•    662-76  +  2      562-236  +  12     462-56  +  l" 


TO   DIVIDE   A   FRACTION  BY   A  FRACTION 

194.  In  arithmetic,  to  divide  a  number  by  the  quotient  of 
two  numbers  is  the  same  as  to  divide  by  the  dividend  and 
multiply  the  result  by  the  divisor. 

E.g.  72  +  —  =  72  +  6  =  12 ; 

if  3 

or,  72  -  ^  =  (72  -  18) .  3  =  4  .  3  =  12. 


Likewise,  jUf  =  (§  +  B)  .  7  =  (^)  .  7 


MULTIPLICATION  AND  DIVISION  275 

Since  in  algebra  a  fraction  is  an  indicated  quotient,  to  divide 
a  number  by  an  algebraic  fraction  we  divide  by  the  numerator 
and  multiply  this  result  by  the  denominator. 


y  b     d     \b       J  \b-cl 


ad 
be' 


Hence,  in  algebra,  as  in  arithmetic,  a  number  is  divided  by  a 
fraction  by  inverting  the  fraction  and  multiplying  by  the  new- 
fraction  thus  obtained. 

Thus,  *^*S5Sfx^»«f. 

b      d     b      c      be 

EXERCISES 

Perform  the  following  indicated  divisions,  and  reduce  the 
resulting  fractions  to  their  lowest  terms : 

q3-[-53        a  +  b  g^-6o;-16  .  x*+9x+U 

d2-9b2  '  a  +  3b'  '  x2+4a;-21  :  tf-Sx+W 

x2  +  x-2  .       a3  4-2^  x2-!       .  a?  +  2x-3 

x>-3x    '  x*  +  9x-36'        '  ^-4^-5  '      x>-25 

a2 -11  a -26  ,  a2- 18  a +  65 
a2_3a_18    *    a2-9a  +  18  ' 

e    x2+9xy  +  lSy2  .  x2  +  6xy  +  9y2 
x2  —  9  xy  +  20  y2  xy2  —  ky* 

x2  4-  mx  +  nx  +  mn      a?  —  n2 
x2  —  mx  —  nx  +  mn      x2  —  m? 

3a4-9a8-54a»     .  a34-8a24-15a 
9  a3 -117  a2 +  378  a  '  3a2-33a  +  84# 

ft    a2- 11  a  +  30     a2- 3  a  .         a2 -9 


a3-6a24-9a      a2-25  '  a24-2a-15 


276  LITERAL  FRACTIONS 


10. 


x*  +  x-56        x-2  +  4x-32 


a2-3a  +  2      a2-8a  +  15  .   a2-9a  +  14 
a2-7a  +  12      «2-6a  +  5    '  a2-12a  +  32* 


12    n2-ll?i  +  18x   n2-8n-9    ,  6n-12 
m2  —  7  ?«  —  18     m2  —  5  m  — 14      an  +  a 


13    a2-62          &(a-5)        .       b(a  +  b) 
'     ab2x       a2  +  2a6  +  62  '  a2-2a&  +  &2 


14    a  +  5         a2-&2      .  (a- &)2(a  +  fr)2 
ab        3(a2+62)  '      3  a3?/ +  3  a?/3 


5a54-5ar>  ^-9^  +  8; 

15. 


7a2-56x-63     14  a;2 +  14  a; -1260 


8y2(,y  +  4)(y  +  5)  ,    y(y  +  4)(y  +  8)  . 
'     2*(y  +  6)(y-7)    '  23G,-7)(j/  +  ll) 

17    (c  +  4)(c-4)      (c  +  2)(c-5)  .  (c  +  4)(c-13) 

•  (C_3)(c_2)      (c-5)(c-6)  '  (c-6)(c  +  ll) 

2a26(6-4)(&  +  4)      (&  +  6)(5+8)  ,  a(&  +  8)(&-4) 

•  5(&  +  4)(6  +  6)        (6  +  8)(6-9)  '    (&-9)(6-5) 

19    ^(s-?)2       (s  +  2)(a-3)    ,  x\x-3)(x-  5) 
'     (x  +  2)2        (a>-2)(a>-7)    '    (s-5)(a>-7) 

a^2(c  +  5)(c-4)     (C-8)(c  +  9)  ,  o6(c+-9)(c-l) 

•  (c_4)(c-8)         (c  +  4)(c  +  7)  "     (c  +  7)(c  +  l) 

21^  +  23a;-  20     6  x2- 11  »- 10  .  7a2  +  17a;-12 
"     10^-27^  +  5        3a»+2o;-5    ""   5x>  +  9x-2  ' 


COMPLEX  FRACTIONS  277 

s*  +  6a;  +  9     9a^  +  9a;-10  ,   3a?-7a;-20 
'      9^-4         7a2  +  20a;-3  '  35a?-12a>  +  l" 


23     (a  +  4)2-62x        3as-4         .  ab*  +  4:b2  +  bs 
12  as3  —  16  x*     ay?  4-  4  a2  —  6X2  jc3 


12 ya^2- 18 ya;       3  6s8  4- 7  6a;  .      3&x3  +  76a;2 
15a;2  +  32aj-7      6a^4-15y  '  10a;24-23a;-5' 

ac  +  bc  —  ar  —  br      6  a2  +  23  a  4-  7    .  3  a2  4-  a  4-  &  +  3  afr 
3  a2  —  8  a  —  3        c&  —  r&  +  c2  —  re  '  a&  —  3  6  4-  ac  —  3  c 


COMPLEX   FRACTIONS 

195.    Sometimes  fractions  occur  whose  numerators  or  de- 
nominators, or  both,  contain  fractions. 

1+1  1     +.    1 


E.g.  2    and  x  +  1     x~1. 

*  1      1  _1 1_ 

a  x  —  1      x  +  1 

Such  fractions  are  called  complex  fractions.  A  complex  frac- 
tion is  said  to  be  simplified  when  it  is  reduced  to  an  equal 
fraction  whose  numerator  and  denominator  are  in  the  integral 
form. 

1+1      «+l 

a     a     a     a+1      a  —  1 


Ex.1. 


■a 1      a  _  1         a  a 


a     a     a 


__<x  +  l         a     _a4-l 
a         a—1      a—1 

This  result  may  also   be   obtained  directly  by  multiplying  both 

I     1 

terms  of  the  given  fraction  by  a,  finding  at  once     "*"    . 

a  —  1 


278  LITERAL   FRACTIONS 


1  +  x 


Ex.2. 


x  +  1     x  —  1 


x  —  1     x+1 


X  —  1  + 

x  +  1 

X2- 

1 

x  +  1  — 

x  +  1 

X2- 

2x 
> 

■  1 

.  x2-! 

x2-! 


By  multiplying  the  terms  of  the  given  fraction  by  (x  +  V)(x  —  1) 

■r  1  4-  X  +  1 

we  may  also  get  directly — — — :!—  =  x. 

x  — j~  X  —  X  ~r  x 


Such  fractions  seldom  occur  in  the  problems  of  elementary 
algebra.  A  more  extended  discussion  is  given  in  the  Advanced 
Course. 

EXERCISES 

Reduce  each  of  the  following  complex  fractions  to  its 
simplest  form : 

m  +  n  +  1  a?  — 8  b3 


1.    1±£.  4.  3  7. 


1  m  —  n  —  1  3a  — 2b 


x 


i_l!  i  i  a  +  ft  Jf 

-         ■   S  2  fl. 


1  +  a'  .      a  —  b  1  +  d2 


4- 


2  Z>2 


,  x  tf-tf  H     hd 

3.    —1.  6.        4  9.     c       cD 

a  g-f-y  1  +  £ 

*~2  2  c. 


RATIO  AND  PROPORTION  279 

RATIO  AND  PROPORTION 

196.  Definitions.  A  fraction  is  often  called  a  ratio.  Thus  - 
may  be  read  the  ratio  of  a  to  b,  and  is  also  written  a  :  b. 

The  numerator  is  called  the  antecedent  of  the  ratio,  and  the 
denominator  the  consequent.  The  antecedent  and  consequent 
are  called  the  terms  of  the  ratio. 

An  equation,  each  of  whose  members  is  a  ratio,  is  called 
a  proportion. 

Thus,  -  =  -  is  a  proportion,  and  is  also  written  a  :  b  =  c :  d. 
b      d 

It  is  read  the  ratio  of  a  tob  equals  the  ratio  of  c  to  d,  or  briefly, 
a  is  to  b  as  c  is  to  d. 

The  four  numbers  a,  b,  c,  and  d,  are  said  to  be  in  proportion. 
a  and  d  are  called  the  extremes  of  the  proportion,  and  6  and  c 
the  means. 

IMPORTANT  PROPERTIES  OF  A  PROPORTION 

1.   If,  in  the  proportion  -  =  -,  both  members  of  the  equation 
b      d 

be  multiplied  by  bd,  we  have,  ad  —  be. 

That  is :  If  four  numbers  are  in  proportion,  the  product  of  the 
means  equals  the  product  of  the  extremes. 


2.  Show  that  if  -  =  c ,  then  -  =  -. 

b      d  a     c 

Hint.   Divide  1  =  1  by  the  members  of  the  given  equation. 
This  process  is  called  taking  the  proportion  by  inversion. 

3.  Show  that  if -=-,  then -  =  -• 

b      d  c      d 

Hint.   Multiply  both  members  of  the  given  equation  by  - . 

c 

This  process  is  called  taking  the  proportion  by  alternation. 


280  LITERAL  FRACTIONS 

4.  Show  that  if  a-  =  c-,  then  ?±±  =  c_±*. 

b      d  b  d 

Hint.    Add  1  to  both  members  of  the  given  equation. 

This  process  is  called  taking  the  proportion  by  composition. 

5.  Show  that  if  -  =  -,  then  *I2*»£i±i!. 

b      d  b  d 

Hint.    Subtract  1  from  each  member  of  the  given  equation. 
This  process  is  called  taking  the  proportion  by  division. 

6.  Show  that  if  ?«*,  then  ^±&  =  ^M. 

b      d  a—  b      c  —  d 

Hint.    Divide  the  members  of  the  equation  obtained  under  4  by 
the  members  of  the  one  obtained  under  5. 

7.  Show  that  if  g  =  ^  =  g,  then  a  +  c  +  e  =  g. 

b      d     f  b+d+f     b 

Let  -  =  -  =  -  =  k :  then  a  =  bk,  c  =  dk.  e  =  fk. 
b     d    f      ■  J 

Hence,  a  +  c  +  e  =  bk  +  dk  +fk  =(b  +  d  +/)  k, 
a  +  c  +  e      7,      a     c      e 


and 


b+d+f  b      d     f 


That  is,  If  several  ratios  are  equal,  the  sum  of  the  antecedents 
is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its 
consequent. 


EXERCISES 


a      c 


1.  If  ad  =  be,  show  that-  =  -•  Hint.    Divide  by  bd. 

b      d 

2.  If  ad  =  be,  show  that  -  =  -  • 

c      d 

3.  If  ad  =  be,  show  that  -  =  -  • 

c      a 


RATIO  AND  PROPORTION  281 


4.    If  ad  =  be,  show  that  -  =  -  ■ 
b      a 


5.   If  -  =  -,  show  that — - — =— —  - 
b      d  a  c 


m     T£  a      c      i         ,i    ,  o-  b      c  —  d 

6.    If  -  =  - ,  show  that = 

b      d  a  c 


m         T£   Cb  C  1  i/U      i.    <*  —  b  C  —  d 

7.  If  -=-,  show  that = 

b     d  a+b     c+d 

8.  If  ^  =  C-,  show  that  1+b^a^dL 

b     d  c+d     c—d 


9.   If  -  =  -,  show  that — ! —  =  -• 
b     d  c+d      c 

10.   If  -  =  -,  show  that  — - — =-• 
b     d  b+d     b 


11.   If-  =  -,  show  that 


a  —  b     a 


b     d  c—d      c 


12.   If -  =  -,  show  that  ^=^  =  -- 
b     d'  b-d     b 

13.  Solve  the  equation  -  =  -  for  each  letter  in  terms  of  all 

6     d 

the  others.  If  a  =  3,  6  =  5,  c  =  8,  find  d.  If  b  =  7,  c  =  9, 
d  =  3,  find  a.  If  c  =  13,  d  =  2,  a  =  5,  find  6.  Ifd  =  50,  a  =  3, 
6= -7,  find  c. 

14.  If  -  =  -,  then  x  is  said  to  be  a  fourth  proportional  to  a,  b, 
,  b     x 

and  c. 

Find  a  fourth  proportional  to  3,  5,  and  7  ;  also  to  9,  5,  and  1, 
and  to  3,  —  2,  and  —  5. 


282  LITERAL  FRACTIONS 

15.  If  -  =  -,  then  x  is  called  a  mean  proportional  between 
a  and  b. 

Solve  the  equation  ~  =  ~  for  *  in  terms  of  a  and  &-     Snow 

that  there  are  two  solutions,  each  of  which  is  a  mean  propor- 
tional between  a  and  b. 

Find  two  mean  proportionals  between  4  and  9 ;  also  between 
5  and  125,  and  between  —  4  and  —  36. 

16.  Which  is  the  greater  ratio  „      , —  or  — t— — 

5  +  4d      5+5d 

Hint.  Reduce  the  fractions  to  a  common  denominator  and  com- 
pare numerators,     (d  is  a  positive  number.) 

17.  Which  is  the  greater  ratio,  a  "**    ,  or  a  ~*~  ,t    ♦ 

a  +  86       a  +  10  6 

18.  Which  is  the  greater  ratio,  -  or     ~*~   ,  if  b  and  c  are  posi- 

b       b  +  c 

tive,  and  a  less  than  b  ?  a  equal  to  b  ?  a  greater  than  b  ? 

19.  Find  two  numbers  in  the  ratio  of  3  to  5  whose  sum  is  160. 

20.  Find  two  numbers  in  the  ratio  of  2  to  7  whose  sum  is 
-108. 

21.  Find  two  numbers  in  the  ratio  of  3  to  —  4  whose  sum 
is  -15. 

22.  What  number  added  to  each  of  the  terms  of  the  ratio 
^  makes  it  equal  to  f|  ? 

23.  What  number  must  be  added  to  each  term  of  the  ratio 
Jy  to  make  it  equal  to  the  ratio  f  ? 

24.  What  number  added  to  each  of  the  numbers  3,  5,  7,  10, 
will  make  the  sums  in  proportion,  when  taken  in  the  given 
order? 

25.  Two  numbers  are  in  the  ratio  of  2  to  3,  and  the  sum  of 
their  squares  is  325.     Find  the  numbers. 


EQUATIONS  INVOLVING  FRACTIONS  283 

EQUATIONS   INVOLVING   FRACTIONS 

197.  We  have  already  seen,  §  40,  that  in  solving  an  equa- 
tion involving  fractions  the  first  step  is  to  multiply  both 
members  of  the  equation  by  a  number  which  will  cancel  all 
the  denominators.  Evidently,  any  common  multiple  of  the 
denominators  is  such  a  multiplier.  For  the  sake  of  sim- 
plicity, and  for  reasons  which  are  fully  discussed  in  the 
Advanced  Course,  the  lowest  common  multiple  is  always  used 
for  this  purpose. 

Illustrative  Example.     Solve  the  equation : 

x  .x  —  6_1—  2  a;  _  55  ,^. 

4+_6  8      "~8"  U 

Solution.  The  L.C.M.  of  4,  6,  and  8  is  24.  Multiplying  both 
members  of  the  equation  by  24, 

24*  ,  24(z-6)      24(1  -  2  x)  _  24  •  55  /0. 

~T+       6       ~        8        ~~ 8"  {2) 

ByF,V,  6x  +  4(:r-6)-3(l  -  2x)=  3  •  55  =  165.  (3) 

ByF,  6z  +  4x-24-3  +  6x  =  165.  (4) 

ByF,  A,  16  a;  =  192.  (5) 

By  A  x  =  12.  (6) 

As  in  this  solution,  so  in  general,  each  denominator  is  can- 
celed by  Principle  V,  after  multiplying  by  the  L.C.M.  In 
practice,  however,  equation  (2)  may  be  omitted,  the  canceling 
being  done  mentally,  and  equation  (3)  may  be  written  down 
at  once. 

The  dividing  line  of  the  fraction  acts  like  a  parenthesis  with 
respect  to  the  sign  preceding  the  fraction. 


284  LITERAL  FRACTIONS 

EXERCISES 

Solve  the  following  equations,  and  check  each  solution  by 
substituting  in  the  original  equation : 

4cc  — 5     2x  +  5__x  —  3  .  7a;  +  5      , 
3  5      =       2     +      8 


_    x  +  3     5x  —  3      5z  —  3_  , 


2a?  +  5  =  5a;  +  20      7a:  +  13      15fl  +  3 
3  15  2  4 


3s  +  10     7a:  +  15     5  a:-14=3  a;-12     2 
5  3  7  4 


,3a  +  5,7-3a;      15  —  a;  ,  3  — 11  x 

5.    X-\ -—-  +  — 7> =  — o 1 a 

4a:  2x  2x  4 

.    7aj-2     3a;-24     3z  +  4     5-4a? 

"• — ~ 7i " 


2^_5»     7_a_^  =  45 
3       12       8       2 


2_x  _  Zx     5x  _  11a;  _     n* 

3        4        7    "  12  = 


„    s-3 . 2s+3     5s-3     3s-l     Q 
9-  -2^  +  "^ 6~-  =  ~^         3- 


,-    2t  +  l     £-1  ,  4*-8     3*-l,„ 
10-  "1 8_  +  "l5~  =  ^0"  +  3- 


,„    z  +  3     4»  +  5  .  3  a-5_5a;-7  ,   . 
"'  ~2 7~~  +  ~4  ^2~  +  4, 


EQUATIONS  INVOLVING  FRACTIONS  285 

8fc-6     13fc     21k-12  =  k-2     Uk-3     d 
'       5  2  5  10k  5 

5a;  — 1      2a;  +  3      5a;  +  l_13a:  +  5     1 
'       2  3  4  11 

14    o       ft-15     ft-20     4fe  +  2  =  Q 
2^3  11 

17  +  ra     2m-7 . 5m—  3  .  Q      3 
15. . _ 1 _ — __f-ii__. 

4  3  2  m 

3r  +  4     5r  +  l     8r+4=o     6r 
5  12  10  7  * 

17    9  +  2*/     4y  +  3_27  +  3y     s 
17>"^ 3» 6~  " 

1Q    2 a; +  5     5a;  +  2  ,  4a;-5_3a;-f  4 

z+3     2z-15     5z-ll     11  =Q 
'2  «  8  2' 

OA3a5  +  20.5a;-3     4a>-l      Q 
20.  __  +  — —  =3. 

198.   Illustrative  Example.    Solve  the  following  equation  : 

2a;  — 1         4     _    3a;       „  m 

x-1      x  +  1     a*-l       '  K  ' 

Solution.  The  L.  C.  M.  of  the  denominators  is  x2  —  1.  In  multi- 
plying both  members  of  the  equation  by  x2— 1,  x  —  1  is  canceled  in 
the  first  fraction,  x  +  1  in  the  second,  and  x2  —  1  in  the  third,  giving 

(2x-l)(x+  l)+4(x-l)-  3x  =  2(x2-  1)  (2) 

Solving,  2x2  +  x- l  +  4x-4-3x  =  2x2- 2,  (3) 

2  x  =  3,  (4) 

and  x  =  §.  (5) 

Check  by  substituting  x  =  §   in  (1). 


286  LITERAL   FRACTIONS 


EXERCISES 

Solve  the  following  equations  and  check  each  solution  by 
substituting  in  the  original  equation  : 

-    Sx-1     4s  +  3        x2  27        1 


2    3a;  +  5     2x  +  l_        x  —  1 


x-9        x  +  2      a?-7x-18 


3   3a;-4     4cc-l  ar>+44      _Q[ 

aj  +  5        a?+4a?  +  9aj  +  20- 


4         4-s2  3a;  +  6  =      2a:  +  l 

ic2  — 5ic  — 14       a:+2  ~         a-  — 7 


a;_4       3a-15  3^-114 

5. 


2z-10      2z-6  4a2-32a:+60 

6   6(a;+4)     3(2a?-l)  =  7[ 
a  +  5  a  +  1         2* 


„    3z-4  .  5a-7  9a^-38 

7-  r  +  ; 


a;-4      2a;-2     2^-10^  +  8 


g   a;  +  17      2(a;  +  6)  =      a;-l 
a;  +  5         ar  +  3    '        a:-f-3 


9    a;  +  2     3a;-15  =  3a:-21 
ic  —  5        a;  — 3  x  —  3 

2a-3  .  3a  +  l     4z  +  17 


10. 


-4a;        <c  — 2         a;  — 2 


EQUATIONS  INVOLVING  FRACTIONS  287 

11    3x-2  =  2xi  +  15x  +  2$     2x-\ 
"  23  +  3       2^  +  53  +  3        3  +  1  ' 

23-3       3-8    _  x  +  2 

23  +  2      53  +  2      23  +  2* 


12. 


20^  +  73-3      3a;  +  l=1 
9a^-l  33-1 


732  +  ll3  +  4       3  +  3  ^73  +  11 
6^  +  13a;  +  5      23  +  1       33  +  5 


33  +  1        a;-3  _=  2ar2-10a;  +  12 
5o;-7     23-7      10.-^-493  +  49' 


199.  Sometimes  it  is  best  to  add  fractions  before  multiplying 
by  the  L.  C.  M.  and  in  other  cases  to  multiply  by  the  L.  C.  M. 
of  part  of  the  denominators  first,  and,  after  simplifying,  multiply 
by  the  L.  C.  M.  of  the  remaining  denominators. 


Ex.  1.   Solve  the  equation 

111 


3—2         3  —  1         3  —  4         3  —  3 

Adding  fractions  on  the  right  and  left, 
1  1 


(3-2)(3-l)        (3-4)(3-3) 


(1) 


(2) 


Multiplying  by  L.  CM.,    (3-4)(3-3)=(3-2)(3-l).  (3) 

Hence  4  3  =  10,  (4) 

and  3  =  2\.  (5) 

Check  by  substituting  3  =  2£  in  equation  (1). 


288  LITERAL  FRACTIONS 

Ex.  2.     Solve  the  equation  : 

4t-3     t-2=2t-2 
16  4        5t  +  2 


3. 


8  dx  +  2 


7*  +  3     21t  +  9  =  17<-3     2 
'       5  15         3*  +  ll 


11^-15      33^  +  15^5^  +  5 
10  30  v-5  ' 


x  —  1      x  —  2  _x  —  3     a  — 4 
#—  2      3  — a;     a;  — 4     a;  — 5 


(1) 


Multiplying  by  16,      4  i  -  3 -4 « +  8  =  ^|f=^-  (2) 

of-)-* 

Hence,  ^W"  <3> 

Multiplying  by  5 1  +  2,  25  £  +  10  =  32  £-32.  (4) 

Hence,  7*  =  42,  (5) 

and  I-.6L  (6) 

Check  by  substituting  £  =  6  in  equation  (1). 

EXERCISES 

3x  +  6     9a;  +  3_  x  +  1       * 
5  15         6  x  -8 

7a:+l      14s-22  =  lla;  +  5 
'      12  24  8  a- 28* 

3z  +  4     12a  +  l      5ic-l 


EQUATIONS  INVOLVING  FRACTIONS  289 

?       1  2  1  4 


X 

-1 

2 

X  +  l 

a; 

-2 

4a:  +  l 

X 

X 

-2 
-3 

a; 

X 

-3 
-4 

X  — 
X  — 

i* 

x  —  5 
6  —  x 

X 

9 

-7 

X 

9 

-2 

5 

X  — 

8 

5 

x  +  l 

X 

-1 

+x 

-2 

X  — 

3 

x  —  5 

9. 


10. 

x— 2     3— #     #— 4     a— 6 


SIMULTANEOUS   FRACTIONAL   EQUATIONS 
200.   When   pairs  of  fractional   equations  are  given,   each 
should  be  reduced  to  the  integral  form  before  eliminating,  ex- 
cept in  special  cases  like  those  in  the  second  following  illustra- 
tive example. 

Ex.  1.    Solve  the  equations : 


x  —  y     x  +  y     tf  —  y* 
3  2  -18 


(2) 


\2x-y     x—3y     (2x  —  y)(x-3y) 

From  (1)  by  M,        4  (x  +  y)  +  6(a>  -  y)  =  36.  (3) 

BjF,D,                                        5x-y  =  l8.  (4) 

From  (2)  by  M,    3(x  -  Sy)  -  2 (2  x -y)  =  - 18.  (5) 

VyF,  D,                                        x  +  ly  =  l%.  (6) 

From  (4)  by  M,                       35x-ly  =  125.  (7) 

Adding  (6)  and  (7),                         36  x  =  144.  (8) 

ByZ>,                                                     a  =  4.  (9) 

Substitute  x  =  4  in  (6),                         y  «  2.  (10) 
Check  by  substituting  x  =  4,  y  =  2  in  (1)  and  (2). 


290 


LITERAL  FRACTIONS 


Ex.  2.   Solve  the  equations 


x      y 


20_21 
Las        y 


=  3. 


(1) 

(2) 


In  this  case  it  is  best  to  solve  the  equations  for  -  and  -  in- 
stead of  for  x  and  y. 


From  (1)  by  M,              —  + 

x 

?i  =  14. 

y 

(3) 

Adding  (2)  and  (3), 

21  =  17. 

X 

(4) 

Hence  by  D, 

1_1 

x     2 

(5) 

Substituting   -  =  -  in  (1), 

1     1 

y    3* 

(6) 

From  (5)  and  (6)  by  M, 

x  =  2,y  =  3. 

(7) 

EXERCISES 

Solve  the  following  equations : 

1.  { 


2a?-l      3y-l_ 


—  «y 


oj  +  1        y+1       (aj  +  lXy  +  l)' 

cc  -f-  2       2x-l_  5xy 


I2y-1        y  +  1        (2y-l)(y+l) 


2.  4 


3a5  +  2_:B  +  l 
3^-5     f  —  1* 
3a;-2     3z-l 


2 


y  +  1        y-1       (2,_l)(y  +  l) 


3. 


PROBLEMS  INVOLVING  FRACTIONS 
(3      2 


291 


5-6-  =  2, 
t      v 

V  t 


f-  +  -  =  21, 

x     y 

I-S— la 

a; 


y 


a     o 
3 


ft 


+  1-?  =  24. 

la      & 


6.  < 


!l 


=  -4, 


-6  +  Ll  =  52. 


7.4 


ri2_io_1 


a;  ?/ 

a  6 

a  2/ 

c  ,  d 

U  y 


h 
1. 


9.  < 


10.^ 


r1.1    ia 

-  +  -  =  16, 

a;     j/ 

1  +  1=14, 

y      z 

i+i=i2. 

2  X 


1    ■    1 

-  +  -=«, 

x     y 

-  +  -=&, 
y     z 

-  +  -  =  c 
Iz      a; 


<> 


In  9  first  add  all  three  equations,  and  from  half  the  sum  subtract 
each  equation  separately.    Likewise  in  10. 


PROBLEMS   LEADING   TO  FRACTIONAL   EQUATIONS 

In  solving  the  following  problems  use  one  or  two  unknowns 
as  may  be  found  most  convenient. 

1.  There  are  two  numbers  whose  sum  is  51  such  that  if  the 
greater  is  divided  by  their  difference,  the  quotient  is  3^.  Find 
the  numbers. 

2.  There  are  two  numbers  whose  sum  is  91  such  that  if  the 
greater  is  divided  by  their  difference,  the  quotient  is  7.  Find 
the  numbers. 

3.  There  are  two  numbers  whose  sum  is  s  such  that  if  the 
greater  is  divided  by  their  difference,  the  quotient  is  q.  Find 
an  expression  in  terms  of  s  and  q  representing  each  number. 
Solve  1  and  2  by  substituting  in  the  formula  just  obtained. 

4.  What  number  must  be  subtracted  from  each  term  of  the 
fraction  \\  so  that  the  result  shall  be  equal  to  ^  ? 


292  LITERAL  FRACTIONS 

5.  What  number  must  be  subtracted  from  each  term  of  the 
fraction  f ^  so  that  the  result  shall  be  equal  to  f  ? 

6.  What  number  must  be  subtracted  from  each  term  of  the 

fraction"  -  so  that  the  result  shall  be  equal  to  -  ?     Solve  4  and 
o  d 

5  by  substituting  in  the  formula  obtained  under  6. 

7.  What  number  must  be  added  to  each  term  of  the  fraction 
\  to  obtain  a  fraction  equal  to  \$  ? 

8.  What  number  must  be  added  to  each  term  of  the  fraction 

-  to  obtain  a  fraction  equal  to  —  ? 
b  H  d 

By  means  of  the  formula  thus  obtained,  solve  problem  7,  and  also 
4,  5,  and  6.     (See  work  under  Problem  23,  p.  114.) 

9.  There  are  two  numbers  whose  difference  is  153.  If  their 
sum  is  divided  by  the  smaller,  the  quotient  is  equal  to  f£. 
Find  the  numbers. 

10.  There  are  two  numbers  whose  difference  is  b.  If  their 
sum  is  divided  by  the  smaller,  the  quotient  is  q.  Find  the 
numbers.     Solve  9  by  substituting  in  this  formula. 

11.  Divide  548  into  2  parts,  such  that  7  times  the  first 
shall  exceed  3  times  the  second  by  474. 

12.  There  are  two  numbers  whose  sum  is  48  such  that  3 
times  the  first  is  8  more  than  5  times  the  second.  Find  both 
numbers. 

13.  There  are  two  numbers  whose  sum  is  s  such  that  a  times 
the  first  is  b  more  than  c  times  the  second.     Find  both  numbers. 

14.  What  number  must  be  subtracted  from  each  of  the  num- 
bers 12, 15, 19,  and  25  in  order  that  the  remainders  may  form  a 
proportion  when  taken  in  the  given  order  ? 

15.  What  number  must  be  added  to  each  of  the  numbers  13, 
21, 3,  and  8  so  that  the  sums  shall  be  in  proportion  when  taken 
in  the  order  given  ? 


PROBLEMS  INVOLVING  FRACTIONS  293 

16.  What  number  must  be  added  to  each  of  the  numbers'  a, 
b,  c,  d  so  that  the  sums  shall  be  in  proportion  when  taken  in 
the  given  order  ? 

17.  What  number  must  be  subtracted  from  each  of  the 
numbers  a,  b,  c,  d  so  that  the  remainders  shall  be  in  proportion 
when  taken  in  the  given  order  ? 

Compare  the  results  in  16  and  17  and  explain  the  relation  between 
them.     (See  remark  under  Problem  23,  page  114.) 

Solve  14  and  15  by  substituting  in  the  formulas  obtained  in  16 
and  17. 

18.  There  is  a  number  composed  of  two  digits  whose  sum  is 
11.  If  the  number  is  divided  by  the  difference  between  the 
digits,  the  quotient  is  16f.  Find  the  number,  the  tens'  digit 
being  the  larger. 

19.  There  is  a  number  composed  of  two  digits  whose  sum 
is  s.  If  the  number  is  divided  by  the  difference  between  the 
digits,  the  quotient  is  q.  Find  the  number,  the  tens'  digit  being 
the  larger. 

20.  Illustrative  Problem.  A  can  do  a  piece  of  work  in  8  days, 
B  can  do  it  in  10  days.  In  how  many  days  can  they  do  it 
together  ? 

Since  A  can  do  the  work  in  8  days,  in  one  day  he  can  do  \  of  it, 
and  since  B  can  do  it  in  10  days,  in  one  day  he  can  do  T^  of  it.  If  x 
is  the  number  of  days  required  when  both  work  together,  in  one  day 

they  can  do  -  of  it.     Hence  we  have  the  equation, 


x 


1  +  1  =  1 
8      10      a; 

21.  A  can  do  a  piece  of  work  in  12  days  and  B  can  do 
it  in  9  days.  How  long  will  it  take  both  working  together 
to  do  it  ? 

22.  A  pipe  can  fill  a  cistern  in  11  hours  and  another  in  13 
hours.     How  long  will  it  require  both  pipes  to  fill  it  ? 


294  LITERAL  FB  ACTIONS 

23.  A  can  do  a  piece  of  work  in  a  days  and  B  can  do  it  in  b 
days.     How  long  will  it  take  both  together  to  do  it  ? 

24.  A  cistern  can  be  filled  by  one  pipe  in  20  minutes  and 
emptied  by  another  in  30  minutes.  How  long  will  it  take  to  fill 
the  cistern  when  both  are  running  together  ? 

25.  A  pipe  can  fill  a  cistern  in  12  hours,  another  in  10  hours, 
and  a  third  can  empty  it  in  8  hours.  How  long  will  it  require 
to  fill  the  cistern  when  they  are  all  running  ? 

26.  A  man  can  do  a  piece  of  work  in  18  days,  another  in  21 
days,  a  third  in  24  days,  and  a  fourth  in  10  days.  How  long 
will  it  require  them  when  all  are  working  together  ? 

27.  A  and  B  working  together  can  do  a  piece  of  work  in  12 
days.  B  and  C  working  together  can  do  it  in  13  days,  and  A 
and  C  working  together  can  do  it  in  10  days.  How  long  will 
it  require  each  to  do  it  when  working  alone  ? 

Suggestion  :  Let  a  =  the  fraction  of  the  work  A  can  do  in  one  day, 
b  =  the  fraction  of  the  work  B  can  do  in  one  day,  and  c  =  the  fraction 
of  the  work  C  can  do  in  one  day. 

Then  a  +  b-^,     b  +  c  =  ^,      c  +  a  =  ^. 

28.  A  and  B  working  together  can  do  a  piece  of  work  in  I 
days.  B  and  C  can  do  it  in  m  days  and  C  and  A  can  do  it  in 
n  days.     How  long  will  it  require  each  working  alone  ? 

29.  The  circumference  of  the  rear  wheel  of  a  carriage  is  1.8 
feet  more  than  that  of  the  front  wheel.  In  running  one  mile 
the  front  wheel  makes  48  revolutions  more  than  the  rear  wheel. 
Find  the  circumference  of  each  wheel. 

If  x  is  the  number  of  feet  in  the  circumference  of  the  front  wheel, 

then  — —  is  the  number  of  revolutions  in  going  one  mile. 
x 

30.  The  circumference  of  the  rear  wheel  of  a  carriage  is  1 
foot  more  than  that  of  the  front  wheel.  In  going  one  mile  the 
two  wheels  together  make  920  revolutions.  Find  the  circum- 
ference of  each. 


MISCELLANEOUS  PROBLEMS  295 

31.  The  distance  from  Chicago  to  Minneapolis  is  420  miles. 
By  increasing  the  speed  of  a  certain  train  7  miles  per  hour  the 
running  time  is  decreased  by  2  hours.  Find  the  speed  of  the 
train. 

4°0 
If  r  is  the  original  rate  of  the  train,  then  — —  is  the  running  time. 

r 

32.  The  distance  from  New  York  to  Buffalo  is  442  miles. 
By  decreasing  speed  of  a  fast  freight  8  miles  per  hour  the 
running  time  is  increased  4  hours.  Find  the  speed  of  the 
freight. 

33.  A  motor  boat  goes  10  miles  per  hour  in  still  water.  In 
10  hours  the  boat  goes  42  miles  up  a  river  and  back  again. 
What  is  rate  of  the  current  ? 

34.  A  train  leaving  New  York  over  the  Pennsylvania  Boad 
requires  9  hours  to  overtake  a  train  leaving  Philadelphia  west- 
ward at  the  same  time.  If  the  Philadelphia  train  had  started 
toward  New  York,  they  would  have  met  in  one  hour.  Find  the 
rate  of  each  train,  the  distance  from  New  York  to  Philadelphia 
being  90  miles. 

35.  The  average  of  the  numbers  x,  34,  0,  —  58,  — 19,  0,  —  20, 
and  y  is  12;  while  the  average  of  2x,3y,  — 18,  50,  and  —  30  is 
—  4.     Find  x  and  y. 

36.  The  length  of  a  rectangle  is  8  feet  greater,  and  its  width 
is  4  feet  greater,  than  the  side  of  a  certain  square.  The  sum 
of  the  areas  of  the  square  and  rectangle  is  736  square  feet. 
Find  the  dimensions  of  each. 

37.  The  difference  between  the  areas  of  a  circle  and  its  cir- 
cumscribed square  is  12  square  inches.  Find  the  radius  of 
the  circle.     (See  problem  33,  p.  239.) 

38.  The  difference  between  the  areas  of  a  circle  ard  its 
inscribed  square  is  12  square  inches.  Find  the  radius  of  the 
circle. 


296  LITERAL  FRACTIONS 

39.  The  difference  between  the  areas  of  a  circle  and  the 
regular  inscribed  hexagon  is  12  square  inches.  Find  the  radius 
of  the  circle. 

40.  The  altitude  of  an  equilateral  triangle  is  6.  Find  its 
side  and  also  its  area.  Find  the  side  and  area,  if  the  altitude 
is  h. 

41.  The  radius  of  a  circle  is  3  feet.  Find  the  area  of  the 
regular  circumscribed  hexagon.  Find  the  area  if  the  radius 
is  r  feet. 

42.  The  radius  of  a  circle  is  r.  Find  the  difference  between 
the  areas  of  the  circle  and  the  regular  circumscribed  hexagon. 

43.  The  difference  between  the  areas  of  a  circle  and  the 
regular  circumscribed  hexagon  is  9  square  inches.  Find  the 
radius  of  the  circle. 

44.  A  circle  is  inscribed  in  a  square  and  another  circum- 
scribed about  it.  The  area  of  the  ring  formed  by  the  two  cir- 
cles is  25  square  inches.     How  long  is  the  side  of  each  square? 

45.  A  square  is  inscribed  in  a  circle  and  another  circum- 
scribed about  it.  The  area  of  the  strip  inclosed  by  the  two 
squares  is  25  square  inches.     Find  the  radius  of  the  circle. 

In  the  following  five  problems  solve  the  resulting  literal  equations 
for  each  letter  in  terms  of  the  others,  and  for  each  solution  state  a 
corresponding  problem. 

46.  A  hound  pursuing  a  deer  gains  400  yards  in  25  minutes. 
If  the  deer  runs  1300  yards  a  minute,  how  fast  does  the  hound 
run  ?  If  the  hound  gains  vx  yards  in  t  minutes  and  the  deer 
runs  v2  yards  per  minute,  find  the  speed  of  the  hound. 

47.  A  disabled  steamer  240  knots  from  port  is  making  only 
4  knots  an  hour.  By  wireless  telegraphy  she  signals  a  tug, 
which  comes  out  to  meet  her  at  17  knots  an  hour.  In  how 
long  a  time  will  they  meet  ?  If  the  steamer  is  s  knots  from 
port  and  making  i>j  knots  per  hour,  and  if  the  tug  makes  v2 
knots  per  hour,  find  how  long  before  they  will  meet. 


MISCELLANEOUS  PROBLEMS  297 

48.  A  motor  boat  starts  7|  miles  behind  a  sailboat  and  runs 
11  miles  per  hour  while  the  sailboat  makes  6^  miles  per  hour. 
How  far  apart  will  they  be  after  sailing  1^  hours  ?  If  the 
motor  boat  starts  s  miles  behind  the  sailboat  and  runs  vy  miles 
per  hour,  while  the  sailboat  runs  v2  miles  per  hour,  how  far 
apart  will  they  be  in  t  hours  ? 

49.  An  ocean  liner  making  21  knots  an  hour  leaves  port 
when  a  freight  boat  making  8  knots  an  hour  is  already  1240 
knots  out.  In  how  long  a  time  will  the  two  boats  be  280  knots 
apart  ?  Is  there  more  than  one  such  position  ?  If  the  liner 
makes  v^  knots  per  hour  and  the  freight  boat,  which  is  st  knots 
out,  makes  v2  knots  per  hour,  how  long  before  they  will  be  s2 
knots  apart? 

50.  A  passenger  train  running  45  miles  per  hour  leaves  one 
terminal  of  a  railroad  at  the  same  time  that  a  freight  running 
18  miles  per  hour  leaves  the  other.  If  the  distance  is  500  miles, 
in  how  many  hours  will  they  meet  ?  If  they  meet  in  8  hours, 
how  long  is  the  road  ?  If  the  rates  of  the  trains  are  vx  and  v2 
and  the  road  is  s  miles  long,  find  the  time. 

51.  In  going  1200  yards  the  rear  wheel  of  a  carriage  makes 
60  revolutions  less  than  the  front  wheel.  If  the  circumference 
of  each  wheel  be  increased  by  3  feet,  the  rear  wheel  will  make 
only  40  revolutions  less  than  the  front  wheel.  Find  the  cir- 
cumference of  each  wheel. 


SCIENCE. 

High  School  Physics. 

By  Professor  Henry  S.  Carhart,  of  the  University  of  Michigan,  and 
H.  N.  CHUTE,  of  the  Ann  Arbor  High  School.  New  edition,  thor- 
oughly revised.     i2rao,  cloth,  440  pages.     Price,  $  1.25. 

NO  other  text-book  on  Physics,  published  in  this  country,  has 
ever  enjoyed  the  popularity  or  the  success  that  has,  from 
the  first  year  of  its  publication,  attended  Carhart  and  Chute's 
High  School  Physics.  Throughout  the  country  the  demand  for 
the  book  has  been  far  in  excess  of  that  for  any  other  manual 
covering  the  same  field,  while  in  many  states  it  has  been  used 
more  widely  than  all  its  competitors  combined.  The  new  edition 
of  the  book  is  a  distinct  improvement  on  its  predecessor.  A 
comparison  of  the  two  books  will  show  numerous  changes  in 
details  and  a  smoothing  down  of  the  rougher  spots.  Physics  is 
not  an  easy  subject ;  but  the  authors  have  made  an  honest  effort 
to  relieve  the  difficulties  for  immature  students  without  such  an 
emasculation  of  the  subject  as  to  diminish  its  value  for  either 
discipline  or  scientific  information. 

The  problems  have  all  been  replaced  by  entirely  new  ones,  in 
which  the  purpose  kept  in  view  is  the  concrete  illustration  of 
principles,  with  a  minimum  of  arithmetical  computation.  Atten- 
tion has  been  given  also  to  the  careful  grading  of  the  problems. 
None  of  them  have  been  inserted  as  puzzles  to  test  the  student's 
intellectual  skill,  while,  as  a  whole,  they  are  distinctly  easier  than 
were  those  of  the  preceding  edition. 

The  book  remains,  as  it  was  before,  the  most  attractive  manual 
for  the  study  of  Physics  that  has  been  published.  Its  method 
cannot  be  improved.  The  principles  of  the  science  are  stated  in 
a  simple,  clear,  and  direct  manner ;  their  application  is  illustrated 
by  apt  experiments ;  finally,  numerous  practical  problems  test  the 
pupil's  mastery  of  the  subject. 

Its  arrangement  of  material,  its  completeness  along  the  lines  of 
the  most  recent  developments  of  physical  science,  its  excellent 
woodcuts,  its  wealth  of  problems,  and  the  ease  and  precision  of 
its  style  deserve  and  have  received  unstinted  praise. 

62 


SCIENCE. 

First  Principles  of  Chemistry. 

By  Raymond  B.  Brownlee,  Far  Rockaway  High  School;  Robert 
W.  Fuller,  Stuyvesant  High  School ;  William  J.  Hancock,  Eras- 
mus Hall  High  School;  Michael  D.  Sohon,  Morris  High  School; 
and  Jesse  E.  Whitsit,  DeWitt  Clinton  High  School;  all  of  New 
York  City. 

THIS  manual  includes  a  treatment  of  the  common  elements 
and  their  important  compounds,  together  with  a  full  dis- 
cussion of  the  fundamental  theories  of  chemistry.  These  theo- 
ries are  simply,  clearly,  and  accurately  stated  without  being  buried 
in  a  confusing  mass  of  detail.  The  recent  developments  in  chem- 
ical theory  are  given  due  recognition. 

In  the  development  of  laws  and  hypotheses,  the  historical  order 
is  followed,  the  experimental  facts  in  each  case  being  described 
before  stating  the  discovery  that  resulted  from  them.  Theoret- 
ical topics  are  introduced  as  soon  as  the  pupil  is  able  to  take  in 
their  full  significance ;  this,  on  the  ground  that  to  delay  their 
presentation  is  to  deprive  the  pupil  of  a  very  useful  tool  in  his 
acquisition  of  chemical  ideas. 

This  book  marks  a  step  in  advance  by  its  omission  of  much 
material  that  has  found  a  place  in  other  elementary  manuals 
through  tradition  rather  than  for  its  real  value.  On  the  other 
hand,  a  number  of  the  metallic  elements,  sometimes  neglected,  are 
given  the  thorough  treatment  that  their  industrial  importance 
deserves.  One  of  the  chief  aims  of  the  descriptive  matter  is  to 
show  the  student  the  many  points  of  contact  between  the  life 
about  him  and  the  science  he  is  studying. 

An  important  feature  of  the  book  is  the  brief  summary  and  the 
test  exercises  given  at  the  end  of  each  chapter.  The  summary 
is  a  series  of  pithy  statements  emphasizing  the  essentials  and 
affording  systematic  review. 

This  book  has  been  prepared  by  five  teachers  of  long  experi- 
ence in  both  college  and  secondary  school  work.  Three  years  of 
careful  discussion  and  revision  by  the  authors  have  produced  a 
unified  text-book  especially  planned  for  the  beginner  in  chemistry. 

56 


SCIENCE. 

Practical  Physiography. 

By  Dr.  Harold  Wellman  Fairbanks,   of  Berkeley,  California. 
8vo,  cloth,  570  pages.     403  Illustrations,     Price,  $1.60. 

IN  this  volume  the  author  has  tried  to  work  out  a  practical,  con- 
crete treatment  of  the  subject  of  Physical  Geography.  The 
book  is  intended  as  an  aid  to  study,  not  as  a  compendium  of 
information  ;  consequently  a  description  of  the  world  as  a  whole 
is  omitted.  Attention  is  devoted  specifically  to  the  region  of  the 
United  States,  and  the  typical  examples  afforded  by  it  are  studied 
as  representatives  of  universal  processes. 

The  purely  descriptive  method  has  been  discarded  as  far  as 
practicable,  the  object  being  to  lead  the  student  to  investigate 
and  find  out  for  himself.  Instead  of  being  told  everything,  he  is 
asked  to  use  his  observing  and  reasoning  powers. 

No  separate  chapters  are  devoted  to  the  relation  between 
physical  nature  and  life,  but,  instead,  this  relation  is  brought 
out  in  its  appropriate  place  in  connection  with  each  topic  through- 
out the  book.  Such  an  arrangement,  it  is  believed,  will  make  the 
whole  matter  much  more  vital. 

Another  feature  which  the  author  trusts  will  meet  with  favor 
from  the  practical  teacher  is  the  distribution  of  the  questions  and 
exercises  throughout  each  chapter,  in  close  connection  with  the 
descriptive  portions  to  which  they  refer.  The  placing  of  ques- 
tions and  exercises  by  themselves  at  the  close  of  each  chapter,  as 
is  done  in  many  text-books  on  the  subject,  puts  a  premium  upon 
mere  memorizing  of  the  text  and  the  omission  of  all  practical 
work.  It  is  not  expected,  however,  that  the  pupils  will  be  able 
to  answer  all  the  questions  without  aid  and  direction  from  the 
teacher. 

The  illustrations  are  a  marked  feature  of  the  book.  Photo- 
graphs have  been  used  wherever  possible,  as  they  appeal  with 
much  more  force  to  pupils  of  high  school  age  than  do  diagrams 
or  sketches.     Most    of  the  views   are  from  the  author's  own 

negatives. 

*  68 


SCIENCE. 

Lessons  in  Elementary  Botany  for  Secondary 
Schools. 

By  Professor  Thomas  H.  Macbride,  Iowa  State  University.     i6mo, 
cloth,  244  pages.     Price,  80  cents. 

THIS  is  a  new  edition  of  Professor  Macbride's  well-known 
text-book,  enlarged  and  revised.  This  edition  contains 
ten  new  full-page  plates,  and  while  it  calls  for  no  material 
but  such  as  is  easily  accessible  to  every  teacher,  its  method 
is  nevertheless  in  perfect  harmony  with  the  best  instruc- 
tion of  the  time.  The  student  is  not  asked  to  study  illustra- 
tions or  text ;  he  is  sent  directly  to  the  plants  themselves 
and  shown  how  to  study  these  and  observe  for  himself  the  vari- 
ous problems  of  vegetable  life.  The  plants  passed  in  review  are 
those  which  are  more  or  less  familiar  to  every  one,  and  the  life 
history  of  which  every  child  should  know. 

Austin  C.  Apgar,  State  Normal  School,  Trenton,  N.J.  :  There  are  many 
points  in  the  book  which  please  me.  Not  the  least  of  these  is  that  the 
author  has  not  been  misled  by  the  craze  for  that  "  logical  order  "  which 
begins  with  protoplasm  and,  some  time  or  other,  if  the  subject  be  pursued 
long  enough,  reaches  such  things  as  can  easily  be  found  and  examined. 
He  begins  with  the  known  and  gradually  advances  to  the  unknown, — 
the  only  order  in  which  successful  teaching  can  be  accomplished. 

A  Key  to  Native  Plants. 

To  accompany  Macbride's  Elementary  Botany. 

IN  response  to  a  desire  expressed  by  many  teachers,  Professor 
Macbride  has  prepared  a  Key  to  the  More  Common  Species 
of  Native  and  Cultivated  Plants  occurring  in  the  Northern  United 
States.  The  material  thus  furnished  has  never  before  been 
offered  in  such  convenient  form  for  class-room  work.  Its  use  will 
in  no  way  change  the  original  plan  of  the  Elementary  Botany, 
but  it  offers  many  advantages  for  additional  study. 

The  Key  will  be  furnished  without  charge  to  those  ordering 
the  author's  Botany.     Copies  ordered  without  the  Botany  will  be 

furnished  at  twenty-five  cents  each. 

60 


MA  THE  AT  A  TICS. 


A  Practical  Course  in  Bookkeeping. 

By  Carlos  B.  Ellis,  of  the  Central  High  School,  Springfield,  Mass. 
8vo,  cloth,  coo  pages.     Price,  $0.00. 

THIS  book  is  the  outgrowth  of  long  experience  in  teaching 
and  in  actual  business.  It  is  designed  to  give  pupils  the 
knowledge  of  the  principles  of  bookkeeping  and  of  the  customs 
of  business  that  every  man  needs,  together  with  sufficient  practice 
in  the  application  of  these  principles  and  customs  to  meet  the 
requirements  of  the  best  business  offices. 

A  feature  of  the  book  is  its  logical  development  of  principles 
whereby  the  pupil  makes  a  careful  study  of  the  essential  accounts 
instead  of  merely  memorizing  rules.  The  pupil  who  understands 
the  uses  of  the  principal  accounts  and  the  relations  which  each 
account  sustains  to  the  others,  will  not  need  to  remember  rules 
for  journalizing.  In  a  word,  the  book  is  designed  to  teach  the 
pupil  to  think. 

A  large  amount  of  work  is  provided  for  supplementary  drill, 
and  there  is  abundant  practice  in  the  use  of  business  forms  and 
of  the  variations  in  method  adapted  to  special  requirements. 

Introduction  to  Geometry. 

By  William  Schoch,  of  the  Crane  Manual  Training  High  School, 
Chicago.     i2mo,  cloth,  142  pages.     Price,  60  cents. 

THE  aim  of  the  author  is  to  furnish  a  series  of  exercises  and 
problems  that  will  enable  the  pupil  to  pass  easily  from  the 
individual  notions  of  his  old  experience  to  a  firm  knowledge  of 
the  elementary  facts  and  concepts  of  Geometry.  The  pupil  is 
to  convince  himself  of  geometric  truths  through  measuring  and 
drawing,  and  thus  gradually  to  come  into  possession  of  the  mate- 
rial with  which  formal  Geometry  deals.  Along  with  this,  the 
practical  side  of  the  subject  is  taken  into  account,  the  pupil's 
knowledge  being  tested  constantly  by  his  power  to  apply  it. 


61 


MA  THE  MA  TICS. 


Plane  and  Spherical  Trigonometry. 

By  President  ELMER  A.  Lyman,  Michigan  State  Normal  College,  and 
Professor  EDWIN  C.  GODDARD,  University  of  Michigan.  Cloth,  146 
pages.    Price,  90  cents. 

THIS  text-book  aims  to  furnish  sufficient  material  in  analytical 
trigonometry,  and  also  in  the  solution  of  the  triangle,  both 
of  which  have  been  adequately  met  heretofore  by  no  single  book. 
Particular  attention  has  also  been  given  to  the  proofs  of  the 
formulae  for  the  functions  of  a  ±  y8.  Nearly  all  other  text-books 
treat  the  same  line  as  both  positive  and  negative  in  the  same 
discussion,  thus  vitiating  the  proof,  and  in  many  cases  proofs 
are  given  for  acute  angles  and  are  then  supposed  to  be  estab- 
lished without  further  discussion  for  all  angles.  These  diffi- 
culties have  been  avoided  by  so  stating  the  proofs  that  the 
language  applies  to  figures  involving  any  angles  and  proving 
the  general  case  algebraically  to  avoid  drawing  an  indefinite 
number  of  such  figures. 

Computation  Tables. 

By  Lyman  and  Goddard.    Price,  50  cents. 

Plane  and  Spherical  Trigonometry,  with  Com- 
putation Tables. 

By  Lyman  and  Goddard.    Complete  edition.     Cloth,  214  pages. 
Price,  $  1.20. 

Principles  of  Plane  Geometry. 

By  J.  W.  MacDonald,  Agent  of  the  Massachusetts  Board  of  Educa- 
tion.    i6mo,  paper,  70  pages.     Price,  30  cents. 


A  Primary  Algebra. 


By  J.  W.  MacDonald,  Agent  of  the  Massachusetts  Board  of  Educa- 
tion. i6mo,  cloth.  The  Complete  Edition,  218  pages  (containing  the 
Teacher's  Guide,  the  Student's  Manual,  and  Answers  to  Problems). 
Price,  75  cents.  The  Student's  Manual,  cloth,  92  pages.  Price,  30  cents. 
64 


MATHEMATICS. 


Plane  Trigonometry. 

By  Professor  R.  D.  BOHANNAN,  of  the  State  University,  Columbus, 
Ohio.     Cloth,  379  pages.     Price,  $2.50. 

SOME  of  the  features  of  this  book  are  the  use  of  triangles  hav- 
ing variety  of  form,  but  always  such  form  as  they  should 
have  if  they  had  been  actually  measured  in  the  field ;  the  sugges- 
tion of  simple  laboratory  exercises ;  problems  from  related  sub- 
jects, such  as  Physics,  Analytical  Geometry,  and  Surveying. 

With  a  view  to  minimize  the  difficulties  of  the  subject,  the 
author  has  ventured  to  present  the  subjects  in  the  following 
order:  the  sine,  its  inverse  and  reciprocal;  the  cosine,  its  in- 
verse and  reciprocal ;  the  sine  and  the  cosine  family  in  union ;  the 
tangent,  its  inverse  and  reciprocal ;  all  the  functions  in  union. 

A  pamphlet  containing  four-place  mathematical  tables  is  fur- 
nished free  with  the  book. 

Calculus  with  Applications 

By  Ellen  Hayes,  Professor  of  Mathematics  at  Wellesley  College. 
i2mo,  cloth,  170  pages.     Price,  $1.20. 

THIS  book  is  a  reading  lesson  in  applied  mathematics,  intended 
for  persons  who  wish,  without  taking  long  courses  in  mathe- 
matics, to  know  what  the  calculus  is  and  how  to  use  it,  either  as 
applied  to  other  sciences,  or  for  purposes  of  general  culture. 
Nothing  is  included  in  the  book  that  is  not  a  means  to  this 
end.  All  fancy  exercises  are  avoided,  and  the  problems  are  for 
the  most  part  real  ones  from  mechanics  or  astronomy. 

Logarithmic  and  Other  Mathematical  Tables. 

By  William  J.  Hussey,  Professor  of  Astronomy  in  the  Leland  Stan- 
ford Junior  University,  California.    8vo,  cloth,  148  pages.     Price,  $1.00. 

VARIOUS  mechanical  devices  make  this  work  specially  easy  to 
consult ;  and  the  large,  clear,  open  page  enables  one  readily 
to  find  the  numbers  sought.  It  commends  itself  at  once  to  the 
eye,  as  a  piece  of  careful  and  successful  book-making. 

65 


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